How to Find Derivatives Using Limits – Step-by-step

The derivative of a function is defined as the limit, which finds the slope of the tangent line to a function. Therefore, we can find an equation by recalling that the slope of a line is found by dividing the change in y by the change in x. This means that we can use limits to find the derivatives of functions.

Here, we will learn how to find derivatives of functions using limits. We will learn a formula and apply it to solve some practice problems.

CALCULUS
Formula for the derivative of a function using limits

Relevant for

Learning to solve derivatives using limits.

See formula

CALCULUS
Formula for the derivative of a function using limits

Relevant for

Learning to solve derivatives using limits.

See formula

Formula for finding derivatives using limits

To find the derivative of a function using limits, we can use the following formula:

$$f'(x)=\lim _{h \to 0}\frac{f(x+h)-f(x)}{h}$$

This formula has the following important parts:

  • $latexf(x+h)$. This means that we have to evaluate the given function using the input $latex x+h$.
  • $latexf(x)$. This is the original function in terms of x.
  • h in the denominator. This value must remain the same until we evaluate the limit or until it is simplified.
  • $latex \lim _{h \to 0}$ is the limit. Generally, we solve it by substituting 0 for all the h that we find.

Therefore, to apply this formula, we have to start by finding the expression $latex f(x+h)$ in the numerator. Then, we subtract the original function $latex f(x)$.

Generally, we can simplify the h in the denominator with the h in the numerator. Finally, we use $latex h=0$ to solve the limit (this will work in most cases, but not always).


Finding derivatives using limits – Examples with answers

The formula for derivatives using limits is applied to solve the following examples. Each example has its respective solution, but try to solve the problems yourself first.

EXAMPLE 1

Use limits to find the derivative of $latex f(x)=5x$.

The formula for the derivative of a function using limits is:

$$f'(x)=\lim _{h \to 0}\frac{f(x+h)-f(x)}{h}$$

Now, we use the $latex function f(x)=5x$ to rewrite the numerator. Therefore, we have:

$$f'(x)=\lim _{h \to 0}\frac{5(x+h)-5x}{h}$$

$$f'(x)=\lim _{h \to 0}\frac{5x+5h-5x}{h}$$

$$f'(x)=\lim _{h \to 0}\frac{5h}{h}$$

We can simplify the h of the numerator with the h of the denominator

$$f'(x)=\lim _{h \to 0}(5)$$

Since we don’t have any h, we simply remove the limit:

$latex f'(x)=5$

EXAMPLE 2

Find the derivative of $latex f(x)=x^2$ using limits.

We start by writing the formula for derivatives with limits:

$$f'(x)=\lim _{h \to 0}\frac{f(x+h)-f(x)}{h}$$

Now, we rewrite the numerator using $latex f(x)=x^2$. Thus, we have:

$$f'(x)=\lim _{h \to 0}\frac{(x+h)^2-x^2}{h}$$

$$f'(x)=\lim _{h \to 0}\frac{x^2+2hx+h^2-x^2}{h}$$

$$f'(x)=\lim _{h \to 0}\frac{2hx+h^2}{h}$$

We can factor out the h from the numerator and simplify with the denominator:

$$f'(x)=\lim _{h \to 0}\frac{h(2x+h)}{h}$$

$$f'(x)=\lim _{h \to 0}(2x+h)$$

Finally, we solve the limit by substituting $latex h=0$ into the expression:

$latex f'(x)=2x+0$

$latex f'(x)=2x$

EXAMPLE 3

Find the derivative of $latex f(x)=2x^2+3$ using limits.

We have the following formula:

$$f'(x)=\lim _{h \to 0}\frac{f(x+h)-f(x)}{h}$$

Then, we use the function $latex f(x)=2x^2+3$ to rewrite the numerator. Thus, we have:

$$f'(x)=\lim _{h \to 0}\frac{(2(x+h)^2+3)-(2x^2+3)}{h}$$

$$f'(x)=\lim _{h \to 0}\frac{2(x^2+2hx+h^2)+3-2x^2-3}{h}$$

$$f'(x)=\lim _{h \to 0}\frac{2x^2+4hx+2h^2+3-2x^2-3}{h}$$

$$f'(x)=\lim _{h \to 0}\frac{4hx+2h^2}{h}$$

Factoring the h from the numerator and simplifying with the h from the denominator, we get:

$$f'(x)=\lim _{h \to 0}\frac{h(4x+2h)}{h}$$

$$f'(x)=\lim _{h \to 0}(4x+2h)$$

Finally, we solve the limit by substituting $latex h=0$ into the expression:

$latex f'(x)=4x+2(0)$

$latex f'(x)=4x$

EXAMPLE 4

Use limits to find the derivative of $latex f(x)=3x^2+5x$.

We start by writing the formula to find derivatives using limits:

$$f'(x)=\lim _{h \to 0}\frac{f(x+h)-f(x)}{h}$$

Now, we use the function $latex f(x)=3x^2+5x$ on the numerator, and we have:

$$f'(x)=\lim _{h \to 0}\frac{(3(x+h)^2+5(x+h))-(3x^2+5x)}{h}$$

$$f'(x)=\lim _{h \to 0}\frac{3(x^2+2hx+h^2)+5x+5h-3x^2-5x}{h}$$

$$f'(x)=\lim _{h \to 0}\frac{3x^2+6hx+3h^2+5x+5h-3x^2-5x}{h}$$

$$f'(x)=\lim _{h \to 0}\frac{6hx+3h^2+5h}{h}$$

We factor the h of the numerator to simplify with the h of the denominator:

$$f'(x)=\lim _{h \to 0}\frac{h(6x+3h+5)}{h}$$

$$f'(x)=\lim _{h \to 0}(6x+3h+5)$$

Finally, we solve the limit by substituting $latex h=0$ into the expression:

$latex f'(x)=6x+3(0)+5$

$latex f'(x)=6x+5$

EXAMPLE 5

Find the derivative of $latex f(x)=\frac{1}{x}$ using limits.

We have the formula:

$$f'(x)=\lim _{h \to 0}\frac{f(x+h)-f(x)}{h}$$

Using the function $latex f(x)=\frac{1}{x}$ on the numerator, we have:

$$f'(x)=\lim _{h \to 0}\frac{\frac{1}{x+h}-\frac{1}{x}}{h}$$

We can use the common denominator $latex (x+h)x$ to subtract the fractions from the numerator:

$$f'(x)=\lim _{h \to 0}\frac{\frac{x-(x+h)}{x(x+h)}}{h}$$

$$f'(x)=\lim _{h \to 0}\frac{\frac{x-x-h}{x^2+xh)}}{h}$$

$$f'(x)=\lim _{h \to 0}\frac{\frac{-h}{x^2+xh)}}{h}$$

We can rewrite the fraction and simplify the h:

$$f'(x)=\lim _{h \to 0}\frac{-h}{h(x^2+xh)}$$

$$f'(x)=\lim _{h \to 0}\frac{-1}{x^2+xh}$$

Finally, we solve the limit by substituting $latex h=0$ into the expression:

$$ f'(x)=\frac{-1}{x^2+x(0)}$$

$$f'(x)=-\frac{1}{x^2}$$

EXAMPLE 6

Determine the derivative of $latex f(x)=\sqrt{x}$ using limits.

We have the formula:

$$f'(x)=\lim _{h \to 0}\frac{f(x+h)-f(x)}{h}$$

Now, we rewrite the numerator using the function $latex f(x)=\sqrt{x}$. Therefore, we have:

$$f'(x)=\lim _{h \to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}$$

We can multiply both the numerator and the denominator by the conjugate of the numerator:

$$f'(x)=\lim _{h \to 0}\frac{(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x})}{h(\sqrt{x+h}+\sqrt{x})}$$

$$f'(x)=\lim _{h \to 0}\frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}$$

$$f'(x)=\lim _{h \to 0}\frac{h}{h(\sqrt{x+h}+\sqrt{x})}$$

We simplify the h in the numerator with the h in the denominator:

$$f'(x)=\lim _{h \to 0}\frac{1}{\sqrt{x+h}+\sqrt{x}}$$

Finally, we solve the limit by substituting $latex h=0$ into the expression:

$$f'(x)=\lim _{h \to 0}\frac{1}{\sqrt{x+0}+\sqrt{x}}$$

$$f'(x)=\lim _{h \to 0}\frac{1}{\sqrt{x}+\sqrt{x}}$$

$$f'(x)=\lim _{h \to 0}\frac{1}{2\sqrt{x}}$$


Finding derivatives using limits – Practice problems

Use the formula for finding derivatives of functions using limits to solve the following practice problems.

Find the derivative of $latex f(x)=7x$.

Choose an answer






What is the derivative of $latex f(x)=7x^2$?

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Find the derivative of $latex f(x)=4x^2-3x$.

Choose an answer






Find the derivative of $latex f(x)=5x^3$.

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Find the derivative of $latex f(x)=-\frac{3}{x}$.

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See also

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