Calculating Derivatives: Problems and Solutions

Are you working to calculate derivatives in Calculus? Let’s solve some common problems step-by-step so you can learn to solve them routinely for yourself.

Jump down this page to: [Power rule, $x^n$] [Exponential, $e^x$] [Trig derivatives] [Product rule] [Quotient rule] [Chain rule]

CALCULUS SUMMARY: Derivatives and Rules

You can always access our Handy Table of Derivatives and Differentiation Rules via the Key Formulas menu item at the top of every page.

CLICK TO VIEW SUMMARY

$x^n$ Derivatives

$$\frac{d}{dx}\text{(constant)} = 0 \quad \frac{d}{dx} \left(x\right) = 1 $$
$$\frac{d}{dx} \left(x^n\right) = nx^{n-1} $$

Exponential Derivative

\begin{align*}
\frac{d}{dx}\left( e^x \right) &= e^x &&& \frac{d}{dx}\left( a^x \right) &= a^x \ln a \\ \\
\end{align*}

Trigonometric Derivatives

\begin{align*}
\frac{d}{dx}\left(\sin x\right) &= \cos x &&& \frac{d}{dx}\left(\csc x\right) &= -\csc x \cot x \\ \\
\dfrac{d}{dx}\left(\cos x\right) &= -\sin x &&& \frac{d}{dx}\left(\sec x\right) &= \sec x \tan x \\ \\
\dfrac{d}{dx}\left(\tan x\right) &= \sec^2 x &&& \frac{d}{dx}\left(\cot x\right) &= -\csc^2 x
\end{align*}

Notice that a negative sign appears in the derivatives of the co-functions: cosine, cosecant, and cotangent.

Constant Factor Rule

Constants come out in front of the derivative, unaffected:
$$\dfrac{d}{dx}\left[c f(x) \right] = c \dfrac{d}{dx}f(x) $$

For example, $\dfrac{d}{dx}\left(4x^3\right) = 4 \dfrac{d}{dx}\left(x^3 \right) =\, … $

Sum of Functions Rule

The derivative of a sum is the sum of the derivatives:
$$\dfrac{d}{dx} \left[f(x) + g(x) \right] = \dfrac{d}{dx}f(x) + \dfrac{d}{dx}g(x) $$

For example, $\dfrac{d}{dx}\left(x^2 + \cos x \right) = \dfrac{d}{dx}\left( x^2\right) + \dfrac{d}{dx}(\cos x) = \, …$

Product Rule for Derivatives

\begin{align*}
\dfrac{d}{dx}(fg)&= \left(\dfrac{d}{dx}f \right)g + f\left(\dfrac{d}{dx}g \right)\\[8px] &= \Big[\text{ (deriv of the 1st) } \times \text{ (the 2nd) }\Big] + \Big[\text{ (the 1st) } \times \text{ (deriv of the 2nd)}\Big] \end{align*}

IV. Quotient Rule for Derivatives

\begin{align*}
\dfrac{d}{dx}\left(\dfrac{f}{g} \right) &= \dfrac{\left(\dfrac{d}{dx}f \right)g – f\left(\dfrac{d}{dx}g \right)}{g^2} \\[8px] &=\dfrac{{\Big[\text{(deriv of numerator) } \times \text{ (denominator)}\Big] – \Big[\text{ (numerator) } \times \text{ (deriv of denominator)}}\Big]}{\text{all divided by [the denominator, squared]}}
\end{align*}

Many students remember the quotient rule by thinking of the numerator as “hi,” the demoninator as “lo,” the derivative as “d,” and then singing

“lo d-hi minus hi d-lo over lo-lo”

[collapse]

I. Power Rule

\[\bbox[yellow,5px]{\dfrac{d}{dx}\left(x^n\right) = nx^{n-1}}\] For example, $\dfrac{d}{dx}\left(x^3\right) = 3x^2;$ $\dfrac{d}{dx}\left(x^{47}\right) = 47x^{46}.$

Two specific cases you’ll quickly remember:
$$\dfrac{d}{dx}\text{(constant)} = 0$$
$$\dfrac{d}{dx}(x) = 1$$

Power Rule Differentiation Problem #1

Differentiate $f(x) = 2\pi$.

Click to View Calculus Solution

$2\pi$ is just a number: it’s a constant. And the derivative of any constant is 0:
\[ \begin{align*}
\dfrac{d}{dx}(2\pi) &= \dfrac{d}{dx}(\text{constant}) \\[8px] &= 0 \quad \cmark
\end{align*} \]

[collapse]

Power Rule Differentiation Problem #2

Find the derivative of $f(x) = \dfrac{2}{3}x^9$.

Click to View Calculus Solution

We’ll show more detailed steps here than normal, since this is the first time we’re using the Power Rule. Recall that $\dfrac{d}{dx}\left(x^n\right) = nx^{n-1}.$
\[ \begin{align*}
\dfrac{d}{dx}\left( \frac{2}{3}x^9\right) &= \frac{2}{3} \dfrac{d}{dx}\left(x^9 \right) \\[8px] &= \frac{2}{3}\left(9 x^{9-1} \right) \\[8px] &= \frac{2}{3}(9) \left(x^8 \right) \\[8px] &= 6x^8 \quad \cmark
\end{align*} \]

[collapse]

Power Rule Differentiation Problem #3

Calculate the derivative of $f(x) = 2x^3 – 4x^2 + x -33$.

Click to View Calculus Solution

Recall that $\dfrac{d}{dx}\left(x^n\right) = nx^{n-1}.$ We simply go term by term:
\[ \begin{align*}
\dfrac{d}{dx}\left(2x^3 – 4x^2 + x -33 \right) &= \dfrac{d}{dx}\left( 2x^3\right) – \dfrac{d}{dx}\left( 4x^2\right) + \dfrac{d}{dx}(x) – \cancelto{0}{\dfrac{d}{dx}(33)} \\[8px] &= 2\dfrac{d}{dx}\left( x^3\right) – 4\dfrac{d}{dx}\left( x^2\right) + \dfrac{d}{dx}(x) \\[8px] &= 2 \left(3 x^2 \right) – 4 \left(2x^1 \right) + 1 \\[8px] &= 6x^2 -8x + 1 \quad \cmark
\end{align*} \]

[collapse]

Power Rule Differentiation Problem #4

Differentiate $f(x) = \sqrt{x}$.

Recall $\sqrt[n]{x} = x^{1/n}.$

Click to View Calculus Solution

Recall that $\dfrac{d}{dx}\left(x^n\right) = nx^{n-1}.$ The rule also holds for fractional powers:
\[ \begin{align*}
\dfrac{d}{dx}\left(\sqrt{x} \right) &= \dfrac{d}{dx}\left(x^{\frac{1}{2}} \right) \\[8px] &= \frac{1}{2}x^{\left(\frac{1}{2}\, – 1 \right)} \\[8px] &= \frac{1}{2}x^{-1/2} \quad \cmark \\[8px] &= \frac{1}{2}\frac{1}{\sqrt{x}} \quad \cmark
\end{align*} \] Note that the last two lines are completely equivalent, and either would be acceptable as the answer.

[collapse]

Power Rule Differentiation Problem #5

Differentiate $f(x) = \dfrac{5}{x^3}$.

Recall $\dfrac{1}{x^n} = x^{-n}.$

Click to View Calculus Solution

Recall that $\dfrac{d}{dx}\left(x^n\right) = nx^{n-1}.$ The rule also holds for negative powers:
\[ \begin{align*}
\dfrac{d}{dx} \left(\dfrac{5}{x^3} \right) &= 5 \dfrac{d}{dx} \left(x^{-3} \right) \\[8px] &= 5 \left((-3) x^{(-3-1)} \right) \\[8px] &= -15 x^{-4} \quad \cmark \\[8px] &= \frac{-15}{x^4} \quad \cmark
\end{align*} \] Note that the last two lines are completely equivalent, and either would be acceptable as the answer.

[collapse]

Power Rule Differentiation Problem #6

Calculate the derivative of $f(x) = \sqrt[3]{x}\, – \dfrac{1}{\sqrt{x}}$.

Click to View Calculus Solution

Recall that $\dfrac{d}{dx}\left(x^n\right) = nx^{n-1}.$
\[ \begin{align*}
\dfrac{d}{dx}\left(\sqrt[3]{x}\, – \dfrac{1}{\sqrt{x}} \right) &= \dfrac{d}{dx} \left(x^{\frac{1}{3}} \right) – \dfrac{d}{dx} \left(x^{-\frac{1}{2}} \right) \\[8px] &= \left(\frac{1}{3}x^{\frac{1}{3}\, – 1} \right) – \left(-\frac{1}{2} x^{\left(-\frac{1}{2}\, – 1 \right)} \right) \\[8px] &= \frac{1}{3} x^{-\frac{2}{3}} + \frac{1}{2}x^{-\frac{3}{2}} \\[8px] &= \frac{1}{3}\frac{1}{\sqrt[3]{x^2}} + \frac{1}{2} \frac{1}{\sqrt{x^3}} \quad \cmark
\end{align*} \]

[collapse]

Power Rule Differentiation Problem #7

Find the derivative of $f(x) = \sqrt{x}\left(x^2 – 8 + \dfrac{1}{x} \right)$.

Click to View Calculus Solution

We’ll learn the “Product Rule” below, which will give us another way to solve this problem. For now, to use only the Power Rule we must multiply out the terms. Recall that $x^a x^b = x^{(a+b)}.$
\[ \begin{align*}
\sqrt{x}\left(x^2 – 8 + \frac{1}{x} \right) &= x^{\frac{1}{2}}x^2 – 8x^{\frac{1}{2}} + x^{\frac{1}{2}}x^{-1} \\[8px] &= x^{\frac{5}{2}}\, – 8x^{\frac{1}{2}}\, + x^{-\frac{1}{2}}
\end{align*} \] We can now take the derivative:
\[ \begin{align*}
\dfrac{d}{dx} \left(\sqrt{x}\left(x^2 – 8 + \frac{1}{x} \right) \right) &= \dfrac{d}{dx}\left(x^{\frac{5}{2}}\right) – 8\dfrac{d}{dx}\left(x^{\frac{1}{2}} \right) + \dfrac{d}{dx} \left( x^{-\frac{1}{2}} \right) \\[8px] &= \frac{5}{2} x^{\left(\frac{5}{2}\, – 1 \right)} – 8 \left(\frac{1}{2} x^{\left(\frac{1}{2}\, – 1 \right)} \right) + \left(-\frac{1}{2} \right)x^{\left(-\frac{1}{2}\, -1 \right)} \\[8px] &= \frac{5}{2}x^{\frac{3}{2}}\, – 4 x^{-\frac{1}{2}} \,- \frac{1}{2} x^{-\frac{3}{2}} \quad \cmark
\end{align*} \]

[collapse]

Power Rule Differentiation Problem #8

Differentiate $f(x) = \left(2x^2 + 1 \right)^2$.

Click to View Calculus Solution

To use only the Power Rule to find this derivative, we must start by expanding the function so we can proceed term by term:
\[ \begin{align*}
\left(2x^2 + 1 \right)^2 &= (2x^2)^2 + 2(2x^2)(1) + 1^2 \\[8px] &= 4x^4 + 4x^2 + 1
\end{align*} \] We can now take the derivative:
\[ \begin{align*}
\dfrac{d}{dx}\left(2x^2 + 1 \right)^2 &= \dfrac{d}{dx}\left(4x^4 \right) + \dfrac{d}{dx} \left(4x^2 \right) + \cancelto{0}{\dfrac{d}{dx}(1)} \\[8px] &= 4(4)x^3 + 4(2)x \\[8px] &= 16x^3 + 8x \quad \cmark
\end{align*} \]

[collapse]

II. Exponential Function Derivative

$$\bbox[yellow,5px]{\dfrac{d}{dx}e^x = e^x}$$
This one’s easy to remember!

Exponential Differentiation Problem #1

Differentiate $f(x) = e^x + x$.

Click to View Calculus Solution

\[ \begin{align*}
\dfrac{d}{dx}\left(e^x + x \right) &= \dfrac{d}{dx}\left(e^x \right) + \dfrac{d}{dx}(x) \\[8px] &= e^x + 1 \quad \cmark
\end{align*} \]

[collapse]

Exponential Differentiation Problem #2

Calculate the derivative of $f(x) = e^{1 + x}$.

Click to View Calculus Solution

Recall that $a^{n+m} = a^n a^m$. Hence
$$e^{1 + x} = e \cdot e^x $$
And remember that $e$ is just a number — a constant. Hence
\[ \begin{align*}
\dfrac{d}{dx}\left(e^{1 + x} \right) &= \dfrac{d}{dx}\left(e \cdot e^x \right) \\[8px] &= e \dfrac{d}{dx}\left(e^x \right) \\[8px] &= e \left(e^x \right) \quad \cmark \\[8px] &= e^{x+1} \quad \cmark
\end{align*} \] Note that the last two lines are completely equivalent. Either is correct as the answer.

[collapse]

III. Trig Function Derivatives

\[ \bbox[yellow,5px]{
\begin{align*}
\frac{d}{dx}\left(\sin x\right) &= \cos x &&& \frac{d}{dx}\left(\csc x\right) &= -\csc x \cot x \\ \\
\dfrac{d}{dx}\left(\cos x\right) &= -\sin x &&& \frac{d}{dx}\left(\sec x\right) &= \sec x \tan x \\ \\
\dfrac{d}{dx}\left(\tan x\right) &= \sec^2 x &&& \frac{d}{dx}\left(\cot x\right) &= -\csc^2 x
\end{align*}} \] Notice that a negative sign appears in the derivatives of the co-functions: cosine, cosecant, and cotangent.

Trig Differentiation Problem #1

Differentiate $f(x) = \sin x – \cos x$.

Click to View Calculus Solution

Recall from the table that $\dfrac{d}{dx}(\sin x) = \cos x,$ and $\dfrac{d}{dx}(\cos x) = -\sin x.$
\[ \begin{align*}
\dfrac{d}{dx} \left(\sin x – \cos x \right) &= \dfrac{d}{dx}(\sin x) – \dfrac{d}{dx}(\cos x) \\[8px] &= \cos x – (-\sin x) \\[8px] &= \cos x + \sin x \quad \cmark
\end{align*} \]

[collapse]

Trig Differentiation Problem #2

Calculate the derivative of $f(x) = 5x^3 – \tan x$.

Click to View Calculus Solution

Recall from the table that $\dfrac{d}{dx}(\tan x) = \sec^2 x.$
\[ \begin{align*}
\dfrac{d}{dx}\left( 5x^3 – \tan x\right) &= 5\dfrac{d}{dx}\left(x^3 \right) – \dfrac{d}{dx}(\tan x) \\[8px] &= 5(3x^2) – \sec^2 x \\[8px] &= 15x^2 – \sec^2 x \quad \cmark
\end{align*} \]

[collapse]

IV. Product Rule

\[\bbox[yellow,5px]{
\begin{align*}
\dfrac{d}{dx}(fg)&= \left(\dfrac{d}{dx}f \right)g + f\left(\dfrac{d}{dx}g \right)\\[8px] &= [{\small \text{ (deriv of the 1st) } \times \text{ (the 2nd) }}]\, + \,[{\small \text{ (the 1st) } \times \text{ (deriv of the 2nd)}}] \end{align*}}\]

Product Rule Differentiation Problem #1

Differentiate $f(x) = x\sin x.$

Click to View Calculus Solution

Since the function is the product of two separate functions, $x$ and $\sin x$, we must use the Product Rule. Recall that $\dfrac{d}{dx}x = 1,$ and that $\dfrac{d}{dx}\sin x = \cos x.$
\[ \begin{align*}
\dfrac{d}{dx} \left( x\sin x\right)&= \left(\dfrac{d}{dx}x\right)\sin x + x \left( \dfrac{d}{dx}\sin x \right) \\[8px] &= (1)\sin x + x \,(\cos x) \\[8px] &= \sin x + x\cos x \quad \cmark
\end{align*} \]

[collapse]

Product Rule Differentiation Problem #2

Calculate the derivative of $f(x) = \left(e^x +1 \right) \tan x.$

Click to View Calculus Solution

Since the function is the product of two separate functions, $ \left(e^x +1 \right)$ and $\tan x$, we must use the Product Rule. Recall that $\dfrac{d}{dx}\left(e^x + 1 \right) = e^x,$ and that $\dfrac{d}{dx}\tan x = \sec^2 x.$
\[\begin{align*}
\dfrac{d}{dx} \left[\left(e^x +1 \right) \tan x \right] &= \left[\dfrac{d}{dx} \left(e^x +1 \right) \right] \tan x + \left(e^x +1 \right)\left[ \dfrac{d}{dx}\tan x \right] \\[8px] &= \left[ e^x \right]\tan x + \left(e^x +1 \right)\left[ \sec^2 x\right] \\[8px] &= e^x \tan x + \left(e^x +1 \right)\sec^2 x \quad \cmark
\end{align*} \]

[collapse]

V. Quotient Rule

\[ \bbox[yellow,5px]{
\begin{align*}
\dfrac{d}{dx}\left(\dfrac{f}{g} \right) &= \dfrac{\left(\dfrac{d}{dx}f \right)g – f\left(\dfrac{d}{dx}g \right)}{g^2} \\[8px] &=\dfrac{{[{\small \text{(derivative of the numerator) } \times \text{ (the denominator)}]}\\ \quad – \, [{\small \text{ (the numerator) } \times \text{ (derivative of the denominator)}}]}}{{\small \text{all divided by [the denominator, squared]}}}
\end{align*}}\] Many students remember the quotient rule by thinking of the numerator as “hi,” the demoninator as “lo,” the derivative as “d,” and then singing

“lo d-hi minus hi d-lo over lo-lo”

Quotient Rule Differentiation Problem #1

Differentiate $f(x) = \dfrac{\sin x}{x}.$

Click to View Calculus Solution

Since the function is the quotient of two separate functions, $\sin x$ and $x$, we must use the Quotient Rule. Recall that $\dfrac{d}{dx}\sin x = \cos x,$ and $\dfrac{d}{dx}x = 1.$
\[ \begin{align*}
\dfrac{d}{dx}\left( \dfrac{\sin x}{x}\right) &= \frac{\left( \dfrac{d}{dx}\sin x \right)x – \sin x \left( \dfrac{d}{dx}x \right)}{x^2} \\[8px] &= \frac{(\cos x)x – \sin x \,(1)}{x^2} \\[8px] &= \frac{x \cos x – \sin x}{x^2} \quad \cmark
\end{align*} \]

[collapse]

Quotient Rule Differentiation Problem #2

Calculate the derivative of $f(x) = \dfrac{e^x}{x+1}.$

Click to View Calculus Solution

Since the function is the quotient of two separate functions, $e^x$ and $(x+1)$, we must use the Quotient Rule. Recall that $\dfrac{d}{dx}e^x = e^x,$ and $\dfrac{d}{dx}(x+1) = 1.$
\[ \begin{align*}
\dfrac{d}{dx} \left(\frac{e^x}{x+1} \right) &= \frac{\left( \dfrac{d}{dx}e^x \right)(x+1) – e^x \left(\dfrac{d}{dx}(x+1) \right)}{(x+1)^2} \\[8px] &= \frac{\left( e^x \right)(x+1) – e^x(1)}{(x+1)^2} \\[8px] &= \frac{e^x(x+1 -1)}{(x+1)^2} \\[8px] &= \frac{xe^x}{(x+1)^2} \quad \cmark
\end{align*} \]

[collapse]

Quotient Rule Differentiation Problem #3

Find the derivative of $f(x) = \dfrac{3x}{5 – \tan x}.$

Click to View Calculus Solution

Since the function is the quotient of two separate functions, $3x$ and $(5 – \tan x)$, we must use the Quotient Rule. Recall that $\dfrac{d}{dx}(3x) = 3,$ and
$$\dfrac{d}{dx}(5 – \tan x) = \cancelto{0}{\dfrac{d}{dx}(5)} – \dfrac{d}{dx}\tan x = -\sec^2 x$$
Then
\[ \begin{align*}
\dfrac{d}{dx}\left(\frac{3x}{5 – \tan x} \right) &= \frac{\left(\dfrac{d}{dx}(3x) \right)(5 – \tan x) – (3x)\left(\dfrac{d}{dx}(5 – \tan x) \right)}{(5 – \tan x)^2} \\[8px] &= \frac{(3)(5 – \tan x) – (3x)(-\sec^2 x)}{(5 – \tan x)^2} \\[8px] &= \frac{15 – 3\tan x + 3x\sec^2 x}{(5 – \tan x)^2} \quad \cmark
\end{align*} \]

[collapse]

VI. Chain Rule

The Chain Rule is a big topic, so we have a separate page on problems that require the Chain Rule.

Need to use the derivative to find the equation of a tangent line (or the equation of a normal line)? We have a separate page on that topic here.

Have a question, suggestion, or item you’d like us to include? Please let us know in the Comments section below!

Want access to all of our Calculus problems and solutions? Buy full access now — it’s quick and easy!

8 Comments

newest

oldest most voted

Inline Feedbacks

View all comments

Anonymous

4 months ago

Thanks you

Matheno

Editor

Reply to Anonymous

4 months ago

You are very welcome! : )

suhasini

9 months ago

In Power Rule Differentiation Problem #3, please explain why d/dx(33)^0 is striked out?

Matheno

Editor

Reply to suhasini

9 months ago

Thanks for asking! That’s just a convenient way to show that the derivative of that term, the constant 33, is 0: d/dx(33) = 0. Any time you have a value that equals zero in an equation, drawing a slanted arrow that points to 0 can help the reader (your teacher, or whomever) see the reason that you’re not carrying the term on to the next line.

Hope that helps, and please let us know any other questions that you have! : )

Ibrahim

1 year ago

I love this idea , and the solution is very good i mean its easy to understand

Matheno

Editor

Reply to Ibrahim

1 year ago

Thanks for letting us know, Ibrahim. That’s our aim: to make things easy for you to understand and then be able to do yourself!

Pratistha

2 years ago

It was really helpful…..thanks for the given solutions which made understanding the topic easier

Matheno

Editor

Reply to Pratistha

2 years ago

Thanks, Pratistha! We’re glad to know that our solutions made understanding how to calculate derivatives easier for you. : )

Rushi

2 years ago

It’s very much helpful

Matheno

Editor

Reply to Rushi

2 years ago

Thanks for writing to tell us. We’ve very happy to know this was useful to you! : )

<img src="/wp-content/themes/matheno2015/images/logo-matheno.png" alt="Matheno - Learn Well and Excel" height="54" width="160"/>

Jump to solved problems