How to Differentiate by First Principles

Video Lesson: How to Do Differentiation from First Principles

What is Differentiation by First Principles?

Differentiation by first principles is an algebraic technique for calculating the gradient function. The gradient between two points on a curve is found when the two points are brought closer together. The gradient is given by the equation f'(x)=lim_h→0[f(x+h)-(fx)]/h.

Derivation of Differentiation by First Principles Equation

Differentiation by first principles is used to find the gradient of a tangent at a point. The method involves finding the gradient between two points. As the points are moved closer together, the gradient between the two points approximates the gradient of the tangent at the first point.

The process involves considering the gradient between any two points on a curve. The gradient between two points can be written as follows:

$g of r a. d i e n t equals the fraction with numerator y sub 2 minus y sub 1 and denominator x sub 2 minus x sub 1$ , where the two points have the coordinates and .

We will consider two points with a horizontal distance between them of ‘h’.

These points will have 𝑥-coordinates of 𝑥 and 𝑥+h.

Therefore these two points will have y-coordinates of and respectively, since they will lie on the curve .

what is differentiation by first principles

The two points have coordinates and .

Therefore , , and .

Therefore, since $g of r a. d i e n t equals the fraction with numerator y sub 2 minus y sub 1 and denominator x sub 2 minus x sub 1$ , the gradient between the two points can be written as: $g of r a. d i e n t equals the fraction with numerator f of open paren x plus h close paren minus f of x and denominator x plus h minus x$ .

This simplifies to $g of r a. d i e n t equals the fraction with numerator f of open paren x plus h close paren minus f of x and denominator h$ .

This equation tells us the gradient between the two points as shown by the red line in the image above.

Since we wish to find the gradient of the tangent to the curve at the location of the first point, the second point is brought closer to the first point.

This is shown in the image above. We can see in the second graph, the red gradient line is a better approximation to the green tangent gradient as the points are closer together.

The gradient between the two points (red line) will equal the gradient of the tangent at the first point (green line) when the distance between the two points approaches zero.

Mathematically, this is equivalent to the distance between the points ‘h’, tending to zero.

We write this as . This means that we are reducing the value of ‘h’ so that it tends to a value of zero.

We add this limit to our gradient equation and write the gradient in function notation as to obtain the first principles equation.

$g of r a. d i e n t equals the fraction with numerator f of open paren x plus h close paren minus f of x and denominator h$ becomes $f prime of x equals lim over h right arrow 0 the fraction with numerator f of open paren x plus h close paren minus f of x and denominator h$ .

proof of differentiation by first principles

Differentiation by First Principles Formula

The differentiation by first principles formula is f'(x)=lim_h→0[f(x+h)-(fx)]/h. For any function f(x), find f(x+h) by replacing x with x+h and substitute f(x+h) and f(x) into the formula. Simplify the numerator and divide all terms by h. Finally evaluate the lim_h→0 by substituting h = 0. The result is the gradient function of f(x).

$f prime of x equals lim over h right arrow 0 the fraction with numerator f of open paren x plus h close paren minus f of x and denominator h$
The formula for differentiation by first principles

In the formula for differentiation by first principles:

f(𝑥) is the function
f(𝑥+h) is found by substituting 𝑥 with 𝑥+h in f(𝑥)
h is the distance between the two points

How to Do Differentiation by First Principles

To do differentiation by first principles:

Find f(x+h) by substituting x with x+h in the f(x) equation.

Substitute f(x+h) and f(x) into the first principles equation.

Simplify the numerator.

Divide all terms by h.

Substituting h=0 to evaluate the limit.

Example: Differentiate f(𝑥) = 2𝑥 + 5 using first principles

We already know that .

Step 1. Find f(𝑥+h) by substituting 𝑥 with 𝑥+h in the f(𝑥) equation

becomes when is replaced with .

Therefore .

Expanding the bracket, .

step 1 of the differentiation by first principles method

Step 2. Substitute f(𝑥+h) and f(𝑥) into the first principles equation

The first principles equation is $f prime of x equals lim over h right arrow 0 the fraction with numerator f of open paren x plus h close paren minus f of x and denominator h$ .

and .

Substituting these in, we obtain: $f prime of x equals lim over h right arrow 0 the fraction with numerator open paren 2 x plus 2 h plus 5 close paren minus open paren 2 x plus 5 close paren and denominator h$ .

step 2 of differentiation by first principles

Step 3. Simplify the numerator

Expanding the brackets on the top of the numerator, $f prime of x equals lim over h right arrow 0 the fraction with numerator open paren 2 x plus 2 h plus 5 close paren minus open paren 2 x plus 5 close paren and denominator h$ becomes $f prime of x equals lim over h right arrow 0 the fraction with numerator 2 x plus 2 h plus 5 minus 2 x minus 5 and denominator h$ .

Notice that the negative sign in front of the (2x+5) results in -2𝑥 and -5.

step 3 of differentiation by first principles

Then the cancels with and +5 cancels with -5.

We are left with .

Step 4. Divide all terms by h

Within there is an h term on the numerator and denominator.

Dividing both the numerator and denominator by h, the equation simplifies to .

step 4 of differentiation by first principles

Step 5. Substitute h=0 to evaluate the limit

To evaluate the limit in , simply substitute h=0 into any h terms after the limit.

After the limit we simply have a 2. There are no h terms in this example.

Therefore nothing changes for us in this step this time.

Therefore .

how to differentiate f(x)=2x+5 using first principles

This means that the gradient at all points along is always equal to 2.

Differentiation by First Principles of a Quadratic

Differentiate using first principles.

Step 1. Find f(𝑥+h) by substituting 𝑥 with 𝑥+h in the f(𝑥) equation

Every 𝑥 term is replaced with 𝑥+h. Substitute the whole of 𝑥+h in brackets.

becomes .

Expanding

and so, .

Step 2. Substitute f(𝑥+h) and f(𝑥) into the first principles equation

and .

$f prime of x equals lim over h right arrow 0 the fraction with numerator f of open paren x plus h close paren minus f of x and denominator h$ becomes $f prime of x equals lim over h right arrow 0 the fraction with numerator x squared plus 2 h of x plus h squared minus open paren x squared minus x close paren and denominator h$ .

step 1 of the first principles method with a quadratic

Step 3. Simplify the numerator

Expanding, .

Therefore $f prime of x equals lim over h right arrow 0 of the fraction with numerator x squared plus 2 h of x plus h squared minus x minus h minus open paren x squared minus x close paren and denominator h$ becomes $f prime of x equals lim over h right arrow 0 of the fraction with numerator x squared plus 2 h of x plus h squared minus x minus h minus x squared plus x and denominator h$ .

Now the cancels with and cancels with .

The equation becomes $f prime of x equals lim over h right arrow 0 of the fraction with numerator 2 h of x plus h squared minus h and denominator h$ .

working out of differentiating a quadratic by first principles

Step 4. Divide all terms by h

All h terms on the numerator and on the denominator must be divided by h.

$f prime of x equals lim over h right arrow 0 of the fraction with numerator 2 h of x plus h squared minus h and denominator h$ becomes .

Step 5. Substitute h=0 to evaluate the limit

To evaluate the limit, substitute h=0.

becomes .

differentiation by first principles of a quadratic

Differentiation by First Principles of a Cubic

Differentiate f(𝑥)=2𝑥³ by first principles.

Step 1. Find f(𝑥+h) by substituting 𝑥 with 𝑥+h in the f(𝑥) equation

To find , substitute with to get .

finding f(x+h) for a cubic in first principles

Now via the binomial expansion .

Therefore .

Step 2. Substitute f(𝑥+h) and f(𝑥) into the first principles equation

$f prime of x equals lim over h right arrow 0 the fraction with numerator f of open paren x plus h close paren minus f of x and denominator h$ becomes $f prime of x equals lim over h right arrow 0 the fraction with numerator 2 x cubed plus 6 x squared h plus 6 x h squared plus 2 h cubed minus 2 x cubed and denominator h$ .

Step 3. Simplify the numerator

The 2𝑥³ cancels with the -2𝑥³ on the numerator.

$f prime of x equals lim over h right arrow 0 the fraction with numerator 2 x cubed plus 6 x squared h plus 6 x h squared plus 2 h cubed minus 2 x cubed and denominator h$ becomes $f prime of x equals lim over h right arrow 0 the fraction with numerator 6 x squared h plus 6 x h squared plus 2 h cubed and denominator h$ .

working out of differentiating a cubic by first principles

Step 4. Divide all terms by h

Dividing all terms by h, $f prime of x equals lim over h right arrow 0 the fraction with numerator 6 x squared h plus 6 x h squared plus 2 h cubed and denominator h$ becomes .

Step 5. Substitute h=0 to evaluate the limit

Finally, substituting all h terms with h=0, becomes .

differentiating 2x^3 with first principles method

Therefore the gradient function of is .

Differentiation by First Principles with Rational Functions

To differentiate a function of 𝑥 with a negative power using the first principles method, write the function as a fraction with a positive power. For example, 𝑥^-1 can be written as ¹/_𝑥.

Differentiate using first principles.

Step 1. Find f(𝑥+h) by substituting 𝑥 with 𝑥+h in the f(𝑥) equation

If then $f of open paren x plus h close paren equals the fraction with numerator 1 and denominator x plus h$ .

Step 2. Substitute f(𝑥+h) and f(𝑥) into the first principles equation

The first principles equation of $f prime of x equals lim over h right arrow 0 the fraction with numerator f of open paren x plus h close paren minus f of x and denominator h$ becomes $f prime of x equals lim over h right arrow 0 the fraction with numerator the fraction with numerator 1 and denominator x plus h minus 1 over x and denominator h$ .

Step 3. Simplify the numerator

The numerator of the fraction contains the subtraction of two fractions.

$the fraction with numerator 1 and denominator x plus h minus 1 over x$ can be written as one fraction with a common denominator of . Simply multiply the numerator and denominator of $the fraction with numerator 1 and denominator x plus h$ by and multiply the numerator and denominator of by .

The fraction subtraction can be written as $x over x times open paren x plus h close paren minus the fraction with numerator open paren x plus h close paren and denominator x times open paren x plus h close paren$ , which can be written as one fraction as $the fraction with numerator x minus open paren x plus h close paren and denominator x times open paren x plus h close paren$ .

differentiation by first principles with fractions

To differentiate a fraction using first principles, combine the fractions formed by into one fraction. This fraction is then within the numerator of the overall fraction in the first principles equation.

To simplify this further, multiply the numerator and denominator of the overall fraction by the denominator of the fraction on the numerator of the overall fraction.

In this example, $f prime of x equals lim over h right arrow 0 of the fraction with numerator the fraction with numerator x minus open paren x plus h close paren and denominator x times open paren x plus h close paren and denominator h$ can be simplified by multiplying by .

$1 lines Line 1: lim over h right arrow 0 the fraction with numerator the fraction with numerator x minus open paren x plus h close paren and denominator x times open paren x plus h close paren and denominator h times x times open paren x plus h close paren over x times open paren x plus h close paren equals lim over h right arrow 0 the fraction with numerator open paren x minus open paren x plus h close paren close paren and denominator h of x times open paren x plus h close paren$ and so, $1 lines Line 1: f prime of x equals lim over h right arrow 0 the fraction with numerator x minus open paren x plus h close paren and denominator h of x times open paren x plus h close paren$ .

Finally, the numerator can simplify further to $1 lines Line 1: f prime of x equals lim over h right arrow 0 the fraction with numerator negative h and denominator h of x times open paren x plus h close paren$ .

differentiate fractions by first principles

Step 4. Divide all terms by h

The numerator and denominator of $1 lines Line 1: f prime of x equals lim over h right arrow 0 the fraction with numerator negative h and denominator h of x times open paren x plus h close paren$ both contain an h term.

Dividing both the numerator and denominator by h, we obtain .

simplifying a fraction when differentiating by first principles

Step 5. Substitute h=0 to evaluate the limit

Finally, to evaluate the limit of , substitute h=0 to obtain .

Simplified, the derivative is $1 lines Line 1: f prime of x equals the fraction with numerator negative 1 and denominator x squared$ .

Differentiate by First Principles of the Square Root of 𝑥

To differentiate a square root function using first principles, multiply the numerator and denominator of the fraction formed by the conjugate of the numerator. This simplifies the equation by removing the square root.

Differentiate .

Step 1. Find f(𝑥+h) by substituting 𝑥 with 𝑥+h in the f(𝑥) equation

If then .

Step 2. Substitute f(𝑥+h) and f(𝑥) into the first principles equation

The first principles equation of $f prime of x equals lim over h right arrow 0 the fraction with numerator f of open paren x plus h close paren minus f of x and denominator h$ becomes $f prime of x equals lim over h right arrow 0 the fraction with numerator the square root of x plus h minus the square root of x and denominator h$ .

Step 3. Simplify the numerator

To simplify the numerator, we multiply the numerator and denominator by the conjugate of the numerator. That is, the negative sign in the numerator replaced with an addition sign. The conjugate of is .

Multiplying . This then simplifies to h.

The full working out for this can be seen as follows.

$the fraction with numerator the square root of x plus h minus the square root of x and denominator h times the fraction with numerator the square root of x plus h plus the square root of x and denominator the square root of x plus h plus the square root of x equals the fraction with numerator open paren the square root of x plus h minus the square root of x close paren times open paren the square root of x plus h plus the square root of x close paren and denominator h of open paren the square root of x plus h plus the square root of x close paren equals the fraction with numerator open paren x plus h close paren plus the square root of x the square root of x plus h minus the square root of x the square root of x plus h minus x and denominator h of open paren the square root of x plus h plus the square root of x close paren equals the fraction with numerator open paren x plus h close paren minus x and denominator h of open paren the square root of x plus h plus the square root of x close paren equals the fraction with numerator h and denominator h of open paren the square root of x plus h plus the square root of x close paren$

differentiating square root x by first principles

Therefore, $f prime of x equals lim over h right arrow 0 the fraction with numerator the square root of x plus h minus the square root of x and denominator h equals lim over h right arrow 0 the fraction with numerator open paren x plus h close paren minus x and denominator h of open paren the square root of x plus h plus the square root of x close paren equals lim over h right arrow 0 the fraction with numerator h and denominator h of open paren the square root of x plus h plus the square root of x close paren$ .

Step 4. Divide all terms by h

In $f prime of x equals lim over h right arrow 0 the fraction with numerator h and denominator h of open paren the square root of x plus h plus the square root of x close paren$ , there is an h term in both the numerator and denominator of the fraction,

Dividing by h, this becomes $f prime of x equals lim over h right arrow 0 the fraction with numerator 1 and denominator the square root of x plus h plus the square root of x$ .

Step 5. Substitute h=0 to evaluate the limit

To evaluate the limit of $f prime of x equals lim over h right arrow 0 the fraction with numerator 1 and denominator the square root of x plus h plus the square root of x$ , substitute h=0 to obtain $f prime of x equals the fraction with numerator 1 and denominator the square root of x plus the square root of x$ .

This simplifies to .

differentiate square root x from first principles

Differentiate by First Principles: f(x) = 1/√x

To differentiate a function with both a fraction and a square root using first principles, a combination of the techniques for differentiating fractions and square roots must be used. That is, combine the two fractions into one fraction first and then multiply through by the conjugate.

Differentiate $f of x equals the fraction with numerator 1 and denominator the square root of x$ using first principles.

Step 1. Find f(𝑥+h) by substituting 𝑥 with 𝑥+h in the f(𝑥) equation

Since $f of x equals the fraction with numerator 1 and denominator the square root of x$ then $f of open paren x plus h close paren equals the fraction with numerator 1 and denominator the square root of x plus h$ .

Step 2. Substitute f(𝑥+h) and f(𝑥) into the first principles equation

Step 3. Simplify the numerator

The two fractions can be combined into one by finding a common denominator of .

The derivative becomes $f prime of x equals lim over h right arrow 0 the fraction with numerator the fraction with numerator the square root of x and denominator the square root of x the square root of x plus h minus the fraction with numerator the square root of x plus h and denominator the square root of x the square root of x plus h and denominator h$ which can be written as $f prime of x equals lim over h right arrow 0 the fraction with numerator the fraction with numerator open paren the square root of x minus the square root of x plus h close paren and denominator the square root of x the square root of x plus h and denominator h$ .

To simplify $f prime of x equals lim over h right arrow 0 the fraction with numerator the fraction with numerator the square root of x minus the square root of x plus h and denominator the square root of x the square root of x plus h and denominator h$ further, multiply by to obtain $f prime of x equals lim over h right arrow 0 the fraction with numerator the square root of x minus the square root of x plus h and denominator h of open paren the square root of x the square root of x plus h close paren$ .

Square roots must be removed on the numerator by multiplying by the conjugate.

The conjugate of is .

$the fraction with numerator the square root of x minus the square root of x plus h and denominator h of open paren the square root of x the square root of x plus h close paren times the fraction with numerator the square root of x plus the square root of x plus h and denominator the square root of x plus the square root of x plus h equals the fraction with numerator x minus open paren x plus h close paren and denominator h of open paren the square root of x the square root of x plus h close paren times open paren the square root of x plus the square root of x plus h close paren$ .

This simplifies the derivative to .

derivative of 1 over square root x by first principles

Step 4. Divide all terms by h

contains an h term on the numerator and denominator.

This simplifies to .

Step 5. Substitute h=0 to evaluate the limit

Substituting h=0, the limit of becomes .

This simplifies to , which can be written as .

differentiate 1 over root x by first principles

Differentiate by First Principles: sin(x)

Differentiating the sine function using first principles involves the compound angle formula for sine and the small angle approximations.

Differentiate .

Step 1. Find f(𝑥+h) by substituting 𝑥 with 𝑥+h in the f(𝑥) equation

Since , .

Step 2. Substitute f(𝑥+h) and f(𝑥) into the first principles equation

The first principles equation of $f prime of x equals lim over h right arrow 0 the fraction with numerator f of open paren x plus h close paren minus f of x and denominator h$ becomes $f prime of x equals lim over h right arrow 0 the fraction with numerator the sine of open paren x plus h close paren minus sine x and denominator h$ .

Step 3. Simplify the numerator

To simplify the numerator, the compound angle formula for sine must be used.

Since , we can write .

The first principles formula becomes $f prime of x equals lim over h right arrow 0 the fraction with numerator open paren sine x times hyperbolic cosine plus cosine x times hyperbolic sine minus sine x close paren and denominator h$ .

compound angle formula sine first principles

To simplify the numerator further, the small angle approximations must be used.

As , the value of h tends to zero.

The small angle approximations tell us that .

Therefore we can replace sin(h) with h and cos(h) with 1.

$f prime of x equals lim over h right arrow 0 the fraction with numerator open paren sine x times hyperbolic cosine plus cosine x times hyperbolic sine minus sine x close paren and denominator h$ becomes $f prime of x equals lim over h right arrow 0 the fraction with numerator sine x times 1 plus cosine x times h minus sine x and denominator h$ .

This simplifies to $f prime of x equals lim over h right arrow 0 the fraction with numerator sine x plus cosine x times h minus sine x and denominator h$ .

This simplifies further to $f prime of x equals lim over h right arrow 0 the fraction with numerator cosine x times h and denominator h$

small angle approximations for the derivative of sinx from first principles

Step 4. Divide all terms by h

Now $f prime of x equals lim over h right arrow 0 the fraction with numerator cosine x times h and denominator h$ simplifies to .

Step 5. Substitute h=0 to evaluate the limit

There are no h terms remaining in and so, substituting h=0, this becomes .

differentiating sinx from first principles

Differentiate by First Principles: cos(x)

Differentiate from first principles.

Step 1. Find f(𝑥+h) by substituting 𝑥 with 𝑥+h in the f(𝑥) equation

Since , .

Step 2. Substitute f(𝑥+h) and f(𝑥) into the first principles equation

The first principles equation of $f prime of x equals lim over h right arrow 0 the fraction with numerator f of open paren x plus h close paren minus f of x and denominator h$ becomes $f prime of x equals lim over h right arrow 0 the fraction with numerator the cosine of open paren x plus h close paren minus cosine x and denominator h$ .

Step 3. Simplify the numerator

To simplify the numerator, the compound angle formula for cosine must be used.

Since , we can write .

The first principles formula becomes $f prime of x equals lim over h right arrow 0 the fraction with numerator cosine x times hyperbolic cosine minus sine x times hyperbolic sine minus cosine x and denominator h$ .

compound angle formula cosine first principles

To simplify the numerator further, the small angle approximations must be used.

As , the value of h tends to zero.

The small angle approximations tell us that .

Therefore we can replace sin(h) with h and cos(h) with 1.

$f prime of x equals lim over h right arrow 0 the fraction with numerator cosine x times hyperbolic cosine minus sine x times hyperbolic sine minus cosine x and denominator h$ becomes $f prime of x equals lim over h right arrow 0 the fraction with numerator cosine x times 1 minus sine x times h minus cosine x and denominator h$ which simplifies to $f prime of x equals lim over h right arrow 0 the fraction with numerator cosine x minus sine x times h minus cosine x and denominator h$ .

The cosx and -cosx terms cancel so this simplifies further to $f prime of x equals lim over h right arrow 0 the fraction with numerator negative sine x times h and denominator h$ .

small angle approximations for differentiating cosine from first principles

Step 4. Divide all terms by h

Now $f prime of x equals lim over h right arrow 0 the fraction with numerator negative sine x times h and denominator h$ simplifies to .

Step 5. Substitute h=0 to evaluate the limit

There are no h terms remaining in and so, substituting h=0, this becomes .

differentiate cosx from first principles

Differentiate by First Principles: ln(x)

Differentiating ln(x) from first principles requires the use of log laws to simplify the equation to one log. Then the limit definition of e^x is used to simplify the equation to ¹/_x.

Differentiate .

Since , .

The first principles equation of $f prime of x equals lim over h right arrow 0 the fraction with numerator f of open paren x plus h close paren minus f of x and denominator h$ becomes $f prime of x equals lim over h right arrow 0 the fraction with numerator l n of open paren x plus h close paren minus l n of x and denominator h$ .

This can be written as .

Using log laws, $f prime of x equals lim over h right arrow 0 1 over h times open bracket l n of open paren the fraction with numerator x plus h and denominator x close paren close bracket$ .

Simplifying the fraction inside the natural logarithm, this becomes .

simpliying lnx when differentiating from first principles

To simplify this further, the limit definition for Euler’s number is used.

Euler’s number e is defined as .

Therefore .

Making the substitution of and by rearrangement of this, , we obtain .

Now replacing with , we can see that .

We had previously seen that from first principles, .

Using log laws, this can be written as .

This can be written as by bringing the logarithm in front of the limit.

Now using the definition of , the derivative becomes .

Finally, since the natural logarithm and the exponential are inverse functions, we obtain .

differentiating lnx from first principles

Differentiate by First Principles: e^x

To differentiate e^x using first principles, a limit definition of lim_h→0[e^h-1]/h = 1 is used.

Differentiate .

Since , .

The first principles equation of $f prime of x equals lim over h right arrow 0 the fraction with numerator f of open paren x plus h close paren minus f of x and denominator h$ becomes $f prime of x equals lim over h right arrow 0 the fraction with numerator e raised to the x plus h power minus e to the x-th power and denominator h$ .

This can be written as $f prime of x equals lim over h right arrow 0 the fraction with numerator e to the x-th power e raised to the h power minus e to the x-th power and denominator h$ or .

Since, it involves no h terms, we can bring e^x outside of the limit to write this as $f prime of x equals e to the x-th power times lim over h right arrow 0 the fraction with numerator open paren e raised to the h power minus 1 close paren and denominator h$ .

To simplify this further, a limit definition of $lim over h right arrow 0 the fraction with numerator open paren e raised to the h power minus 1 close paren and denominator h equals 1$ is used.