How To Find Derivatives

Calculus

How to Find Derivatives in 3 Steps

02.15.2022 • 9 min read

Rachel McLean

Subject Matter Expert

What is a derivative, and how can we calculate it? In this article, we’ll discuss the definition of a derivative and 3 steps to differentiate functions. Then, you can test your knowledge with some examples.

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In This Article

  1. What is a Derivative?

  2. 3 Steps to Find Derivatives

  3. 4 Practice Exercises

What is a Derivative?

Derivatives measure the instantaneous rate of change of a function. When we talk about rates of change, we’re talking about slopes.

The instantaneous rate of change of a function at a point is equal to the slope of the function at that point. When we find the slope of a curve at a single point, we find the slope of the tangent line. The tangent line to a function at a point is a line that just barely touches the function at that point.

So, the derivative of a function at a single point is equal to the slope of the tangent line at that point.

To visualize the tangent line, let’s look at an example where the equation for the tangent line has already been calculated. Consider the function f(x)=ln(x)f(x) = \ln{(x)}, indicated by the blue line.

graph showing how to find Derivatives

Suppose we want to find the derivative of f(x)f(x) at the point (1,0)(1, 0). The tangent line to the curve f(x)f(x) at (1,0)(1, 0) is represented by the red line, f(x)=x1f(x) = x - 1. Notice how this line touches f(x)=ln(x)f(x) = \ln{(x)} at just one point, (1,0)(1, 0).

The tangent line f(x)=x1f(x) = x - 1 is given in the slope-intercept form f(x)=mx+bf(x) = mx +b, where mm is the slope. So, we can easily see that the slope of f(x)=x1f(x) = x - 1 is 1.

This means that the instantaneous rate of change, or derivative, of the function f(x)=ln(x)f(x) = \ln{(x)} at the point (0,1)(0, 1) is 1.

The instantaneous rate of change of ff at aa, or the derivative of ffat aa, is represented by the notation f(a)f’(a). We read aloud the symbol f(a)f’(a) as either “the derivative of ff evaluated at aa” or “ff prime at aa.”

The general derivative function of y=f(x)y = f(x) is usually represented by either f(x)f’(x) or dydx\frac{dy}{dx}. (You can read more about the meaning of dy/dx if needed.) This function tells us the instantaneous rate of change of ff with respect to xx at any point on the curve.

You can watch Dr. Hannah Fry explain more about what a derivative is below.

3 Steps to Find Derivatives

Next, we’ll learn exactly how to find the derivative of a function. So far, we’ve learned that the slope at a point on a curve is called the slope of the tangent line or the instantaneous rate of change.

By contrast, the slope between two separate points on a curve is called the slope of the secant line. This slope value is also referred to as the average rate of change. The average rate of change will help us calculate the derivative of a function.

To find the average rate of change, we divide the change in the output values (y-values) by the change in the input values (x-values). The delta symbol Δx\Delta{x} represents the "change in xx," which is the value that xx is changing by. The average rate of change of the function ff over the interval [a,b][a, b] is:

Average Rate of Change=ΔyΔx=y2y1x2x1=f(b)f(a)ba\text{Average Rate of Change} = \frac{\Delta{y}}{\Delta{x}} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{f(b)-f(a)}{b-a}

By making Δx\Delta{x} approach 0, we can find the instantaneous rate of change. Take a look at the limit definition of a derivative below.

The derivative of ff at xx is equal to the limit of the average rate of change of ff over the interval [x,x+Δx][x, x +\Delta{x}] as Δx\Delta{x} approaches 0. This limit is given by:

f(x)=limΔx0ΔyΔx=limΔx0f(x+Δx)f(x)Δx=Lf’(x) = \mathop {\lim }\limits_{\Delta{x} \to 0}\frac{\Delta{y}}{\Delta{x}}=\mathop{\lim }\limits_{\Delta{x} \to 0} \frac{{f\left( {x + \Delta{x} } \right) - f\left( x\right)}}{\Delta{x} }=L

If this limit exists, then f(x)f(x) is differentiable, and the derivative of the function ff at xx is LL. For a brief review of limits, read What Are Limits? and How to Find Limits.

Here are 3 simple steps to calculating a derivative:

  1. Substitute your function into the limit definition formula.

  2. Simplify as needed.

  3. Evaluate the limit.

Let’s walk through these steps using an example. Suppose we want to find the derivative of f(x)=2x2f(x) = 2x^2.

Step 1

First, we need to substitute our function f(x)=2x2f(x) = 2x^2 into the limit definition of a derivative. Substituting the first term of the limit definition’s numerator correctly can be tricky at first. The key is to simply substitute xx with (x+Δx)(x + \Delta{x}) wherever xx appears in the function.

f(x)=limΔx0f(x+Δx)f(x)Δxf’(x)= \mathop {\lim }\limits_{\Delta{x} \to 0} \frac{{f\left( {x + \Delta{x} } \right) - f\left( x \right)}}{\Delta{x} }
=limΔx02(x+Δx)22x2Δx= \mathop {\lim }\limits_{\Delta{x} \to 0} \frac{2(x + \Delta{x})^2 - 2x^2}{\Delta{x}}

Step 2

Next, we simplify our function as much as we can. First, we’ll expand the term 2(x+Δx)22(x + \Delta{x})^2, and then combine like terms. Then, since Δx\Delta{x} is present in all terms of the numerator and denominator, we can divide by Δx\Delta{x}.

f(x)=limΔx02(x2+2xΔx+Δx2)2x2Δxf’(x)= \mathop {\lim }\limits_{\Delta{x} \to 0} \frac{2(x^2 + 2x\Delta{x} + \Delta{x}^2) - 2x^2}{\Delta{x}}
=limΔx02x2+4xΔx+2Δx22x2Δx= \mathop {\lim }\limits_{\Delta{x} \to 0} \frac{2x^2 + 4x\Delta{x} + 2\Delta{x}^2 - 2x^2}{\Delta{x}}
=limΔx04xΔx+2Δx2Δx= \mathop {\lim }\limits_{\Delta{x} \to 0} \frac{4x\Delta{x} + 2\Delta{x}^2}{\Delta{x}}
=limΔx04x+2Δx= \mathop {\lim }\limits_{\Delta{x} \to 0}4x + 2\Delta{x}

Step 3

Finally, we can evaluate the limit as Δx\Delta{x} approaches 0. Since we’re left with a polynomial function and polynomials are always continuous, we can simply substitute Δx=0\Delta{x} = 0 into the function we’re left with.

f(x)=limΔx04x+2Δxf’(x)= \mathop {\lim }\limits_{\Delta{x} \to 0}4x + 2\Delta{x}
=4x+2(0)= 4x + 2(0)
=4x= 4x

Thus, the general derivative formula of f(x)=2x2f(x) = 2x^2 is 4x4x. If you want to find the derivative at a single point, you can simply plug x=ax = a into f(x)=4xf’(x) = 4x. For example:

  • f(1)=4(1)=4f’(1) = 4(1) = 4. This value represents the slope of the tangent line at x=1x = 1.

  • f(2)=4(2)=8f’(2) = 4(2) = 8. This value represents the slope of the tangent line at x=2x = 2.

  • f(100)=4(100)=400f’(100) = 4(100) = 400. This value represents the slope of the tangent line at x=100x = 100.

Standard Derivative Rules

Now that you’re familiar with the limit definition of a derivative, you can begin to memorize the standard derivative rules below. These derivative rules are derived from the limit definition of a derivative. They allow us to evaluate derivatives much faster.

Here are some of the most common derivative rules to know:

Constant Rule

ddxc=0\frac{d}{dx}c = 0

Power Rule

ddx(xn)=nxn1\frac{d}{dx}(x^n) = nx^{n-1}

Special Case of the Power Rule (where n=1): ddx(x)=1\frac d{dx}(x)=1

Constant Multiple Rule

ddx(cf(x))=cf(x)\frac d{dx}(c\cdot f(x))=c\cdot f'(x)

Chain Rule

ddxf(g(x))=f(g(x))g(x)\frac{d}{dx}f(g(x)) = f’(g(x))g’(x)

Product Rule

ddx[f(x)g(x)]=f(x)g(x)+f(x)g(x)\frac{d}{dx}[f(x) \cdot g(x)] = f’(x) \cdot g(x) + f(x)\cdot g’(x)

Quotient Rule

ddx[f(x)g(x)]=g(x)f(x)f(x)g(x)(g(x))2\frac{d}{dx}[\frac{f(x)}{g(x)}] = \frac{g(x)f’(x)-f(x)g’(x)}{(g(x))^2}

Sum/Difference Rule

ddx[f(x)±g(x)]=f(x)±g(x)\frac{d}{dx}[f(x) \pm g(x)] = f’(x) \pm g’(x)

Trigonometry Rules

ddx(sin(x))=cos(x)\frac{d}{dx}(\sin{(x)}) = \cos{(x)}

ddx(cos(x))=sin(x)\frac{d}{dx}(\cos{(x)}) = -\sin{(x)}

ddx(tan(x))=sec2(x)\frac{d}{dx}(\tan{(x)}) = \sec ^2 (x)

Logarithmic and Exponential Rules

ddx(lnx)=1x\frac{d}{dx} (\ln{x}) = \frac{1}{x}

ddx(ex)=ex\frac{d}{dx}(e^x) = e^x

Dr. Tim Chartier explains more about the Product and Quotient derivative rules.

4 Practice Exercises

Exercise 1

Let f(x)=7x1f(x) = 7x - 1. Using the limit definition of a derivative, find f(x)f’(x).

Solution

Substituting your function into the limit definition can be the hardest step for functions with multiple terms. Remember to double-check your answer, use parentheses where necessary, and distribute negative signs appropriately.

f(x)=limΔx0f(x+Δx)f(x)Δxf’(x)= \mathop {\lim }\limits_{\Delta{x} \to 0} \frac{{f\left( {x + \Delta{x} } \right) - f\left( x \right)}}{\Delta{x}}
=limΔx07(x+Δx)1(7x1)Δx= \mathop {\lim }\limits_{\Delta{x} \to 0} \frac{7(x + \Delta{x}) - 1 - (7x - 1)}{\Delta{x}}
=limΔx07x+7Δx17x+1Δx= \mathop {\lim }\limits_{\Delta{x} \to 0} \frac{7x + 7\Delta{x} - 1 - 7x + 1}{\Delta{x}}
=limΔx07ΔxΔx= \mathop {\lim }\limits_{\Delta{x} \to 0} \frac{7\Delta{x}}{\Delta{x}}
=limΔx07= \mathop {\lim }\limits_{\Delta{x} \to 0} 7
=7=7

So f(x)=7f’(x) = 7.

Exercise 2

Let f(x)=1xf(x) = \frac{1}{x}. Using the limit definition of a derivative, find f(x)f’(x).

Solution

After substituting our function into the limit definition, we’ll need to combine the two fractions in the numerator by finding a common denominator and then multiplying appropriately.

f(x)=limΔx0f(x+Δx)f(x)Δxf’(x)= \mathop {\lim }\limits_{\Delta{x} \to 0} \frac{{f\left( {x + \Delta{x} } \right) - f\left( x \right)}}{\Delta{x}}
=limΔx01x+Δx1xΔx= \mathop {\lim }\limits_{\Delta{x} \to 0} \frac{\frac{1}{x + \Delta{x}} - \frac{1}{x}}{\Delta{x}}
=limΔx0xx(x+Δx)(x+Δx)x(x+Δx)Δx= \mathop {\lim }\limits_{\Delta{x} \to 0} \frac{\frac{x}{x(x + \Delta{x})} - \frac{(x+ \Delta{x})}{x(x+ \Delta{x})}}{\Delta{x}}
=limΔx0xxΔxx(x+Δx)Δx= \mathop {\lim }\limits_{\Delta{x} \to 0} \frac{\frac{x-x-\Delta{x}}{x(x+\Delta{x})}}{\Delta{x}}
=limΔx0Δxx(x+Δx)1Δx= \mathop {\lim }\limits_{\Delta{x} \to 0} \frac{-\Delta{x}}{x(x+ \Delta{x})} \cdot \frac{1}{\Delta{x}}
=limΔx01x(x+Δx)= \mathop {\lim }\limits_{\Delta{x} \to 0} \frac{-1}{x(x+\Delta{x})}
=1x(x+0)= \frac{-1}{x(x+0)}
=1xx= \frac{-1}{x \cdot x}
=1x2= -\frac{1}{x^2}

So, f(x)=1x2f’(x) = -\frac{1}{x^2}.

Exercise 3

Let f(x)=3x2+ecos(x)f(x) = 3x^2 + e^{cos{(x)}}. Using the derivative rules, find f(x)f’(x).

Solution

For this problem, we’ll need to use the Sum Rule. The Sum Rule states that the derivative of a sum of functions is equal to the sum of their derivatives. To find the derivative of each separate function, we can use the Power Rule and the Constant Multiple Rule for the first term, and the Chain Rule, trigonometry rules, and the exponential rule for the second term.

For the first term, we use the Power Rule with n=2n=2, combining that with the Constant Multiple Rule where c=3c=3. Using these, we see that the derivative of the first term is f(x)=32x21=32x=6xf'(x)=3\cdot2x^{2-1}=3\cdot2x=6x.

For the second term, we also have a composition of functions with ecos(x)e^{\cos{(x)}}. The Chain Rule says that the derivative of a composition of functions is found by first taking the derivative of the "outside" function and leaving the “inside” unchanged, and then multiplying by the derivative of the "inside" function.

So, to find the derivative of ecos(x)e^{\cos{(x)}}, we can simply use the exponential rule and then multiply by sin(x)-\sin{(x)}, which is the derivative of the inside function cos(x)\cos{(x)}.

f(x)=6x+ecos(x)(sin(x))f’(x) = 6x + e^{\cos{(x)}}\cdot (-\sin{(x)})
f(x)=6xsin(x)ecos(x)f’(x) = 6x -\sin{(x)}e^{\cos{(x)}}

So, f(x)=6xsin(x)ecos(x)f’(x) = 6x -\sin{(x)}e^{\cos{(x)}}.

Exercise 4

Let f(x)=2xsin(5x)f(x) = 2x\sin{(5x)}. Using the derivative rules, find f(x)f’(x).

Solution

For this problem, we’ll need to use the Product Rule. The product rule states that the derivative of a product of functions is the sum of the first function times the derivative of the second and the second function times the derivative of the first.

The first function in our product is 2x2x. To find the derivative of this we use the special case of the Product Rule with n=1n=1 as well as the Constant Multiple Rule with c=2c=2. Applying these rules, we find that the derivative of 2x2x is

ddx(2x)=21=2\frac d{dx}\left(2x\right)=2\cdot1=2

For the second function in our product, we have a composition of functions with sin(5x)\sin{(5x)}. The Chain Rule says that the derivative of a composition of functions is found by first taking the derivative of the "outside" function and leaving the “inside” unchanged, and then multiplying by the derivative of the "inside" function.

So, to find the derivative of sin(5x)\sin{(5x)}, we can simply use the trigonometry rules for sin and then multiply by 55, which is the derivative of the inside function 5x5x. The derivative of the inside function is found by using the special case of the Power Rule for n=1n=1 and the Constant Multiple Rule.

Plugging the derivatives we found into the Product Rule, we get:

f(x)=2xcos(5x)5+sin(5x)2f’(x) = 2x \cdot \cos{(5x)} \cdot 5 + \sin{(5x)} \cdot 2
f(x)=10xcos(5x)+2sin(5x)f’(x) = 10x\cos{(5x)} + 2\sin{(5x)}

So, f(x)=10xcos(5x)+2sin(5x)f’(x) = 10x\cos{(5x)} + 2\sin{(5x)}.

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