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Calculating Derivatives: Problems and Solutions

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”

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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:
\[ \begin{align*}
\dfrac{d}{dx}(2\pi) &= \dfrac{d}{dx}(\text{constant}) \\[8px] &= 0 \quad \cmark
\end{align*} \]
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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 $\dfrac{d}{dx}\left(x^n\right) = nx^{n-1}.$
\[ \begin{align*}
\dfrac{d}{dx}\left( \frac{2}{3}x^9\right) &= \frac{2}{3} \dfrac{d}{dx}\left(x^9 \right) \\[8px] &= \frac{2}{3}\left(9 x^{9-1} \right) \\[8px] &= \frac{2}{3}(9) \left(x^8 \right) \\[8px] &= 6x^8 \quad \cmark
\end{align*} \]
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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:
\[ \begin{align*}
\dfrac{d}{dx}\left(2x^3 – 4x^2 + x -33 \right) &= \dfrac{d}{dx}\left( 2x^3\right) – \dfrac{d}{dx}\left( 4x^2\right) + \dfrac{d}{dx}(x) – \cancelto{0}{\dfrac{d}{dx}(33)} \\[8px] &= 2\dfrac{d}{dx}\left( x^3\right) – 4\dfrac{d}{dx}\left( x^2\right) + \dfrac{d}{dx}(x) \\[8px] &= 2 \left(3 x^2 \right) – 4 \left(2x^1 \right) + 1 \\[8px] &= 6x^2 -8x + 1 \quad \cmark
\end{align*} \]
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Power Rule Differentiation Problem #4
Differentiate $f(x) = \sqrt{x}$.
Tips iconRecall $\sqrt[n]{x} = x^{1/n}.$
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.
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Power Rule Differentiation Problem #5
Differentiate $f(x) = \dfrac{5}{x^3}$.
Tips iconRecall $\dfrac{1}{x^n} = x^{-n}.$
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.
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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 $\dfrac{d}{dx}\left(x^n\right) = nx^{n-1}.$
\[ \begin{align*}
\dfrac{d}{dx}\left(\sqrt[3]{x}\, – \dfrac{1}{\sqrt{x}} \right) &= \dfrac{d}{dx} \left(x^{\frac{1}{3}} \right) – \dfrac{d}{dx} \left(x^{-\frac{1}{2}} \right) \\[8px] &= \left(\frac{1}{3}x^{\frac{1}{3}\, – 1} \right) – \left(-\frac{1}{2} x^{\left(-\frac{1}{2}\, – 1 \right)} \right) \\[8px] &= \frac{1}{3} x^{-\frac{2}{3}} + \frac{1}{2}x^{-\frac{3}{2}} \\[8px] &= \frac{1}{3}\frac{1}{\sqrt[3]{x^2}} + \frac{1}{2} \frac{1}{\sqrt{x^3}} \quad \cmark
\end{align*} \]
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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:
\[ \begin{align*}
\dfrac{d}{dx} \left(\sqrt{x}\left(x^2 – 8 + \frac{1}{x} \right) \right) &= \dfrac{d}{dx}\left(x^{\frac{5}{2}}\right) – 8\dfrac{d}{dx}\left(x^{\frac{1}{2}} \right) + \dfrac{d}{dx} \left( x^{-\frac{1}{2}} \right) \\[8px] &= \frac{5}{2} x^{\left(\frac{5}{2}\, – 1 \right)} – 8 \left(\frac{1}{2} x^{\left(\frac{1}{2}\, – 1 \right)} \right) + \left(-\frac{1}{2} \right)x^{\left(-\frac{1}{2}\, -1 \right)} \\[8px] &= \frac{5}{2}x^{\frac{3}{2}}\, – 4 x^{-\frac{1}{2}} \,- \frac{1}{2} x^{-\frac{3}{2}} \quad \cmark
\end{align*} \]
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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:
\[ \begin{align*}
\dfrac{d}{dx}\left(2x^2 + 1 \right)^2 &= \dfrac{d}{dx}\left(4x^4 \right) + \dfrac{d}{dx} \left(4x^2 \right) + \cancelto{0}{\dfrac{d}{dx}(1)} \\[8px] &= 4(4)x^3 + 4(2)x \\[8px] &= 16x^3 + 8x \quad \cmark
\end{align*} \]
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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
\[ \begin{align*}
\dfrac{d}{dx}\left(e^x + x \right) &= \dfrac{d}{dx}\left(e^x \right) + \dfrac{d}{dx}(x) \\[8px] &= e^x + 1 \quad \cmark
\end{align*} \]
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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.
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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 $\dfrac{d}{dx}(\cos x) = -\sin x.$
\[ \begin{align*}
\dfrac{d}{dx} \left(\sin x – \cos x \right) &= \dfrac{d}{dx}(\sin x) – \dfrac{d}{dx}(\cos x) \\[8px] &= \cos x – (-\sin x) \\[8px] &= \cos x + \sin x \quad \cmark
\end{align*} \]
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Trig Differentiation Problem #2
Calculate the derivative of $f(x) = 5x^3 – \tan x$.
Click to View Calculus Solution
Recall from the table that $\dfrac{d}{dx}(\tan x) = \sec^2 x.$
\[ \begin{align*}
\dfrac{d}{dx}\left( 5x^3 – \tan x\right) &= 5\dfrac{d}{dx}\left(x^3 \right) – \dfrac{d}{dx}(\tan x) \\[8px] &= 5(3x^2) – \sec^2 x \\[8px] &= 15x^2 – \sec^2 x \quad \cmark
\end{align*} \]
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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 $\dfrac{d}{dx}\sin x = \cos x.$
\[ \begin{align*}
\dfrac{d}{dx} \left( x\sin x\right)&= \left(\dfrac{d}{dx}x\right)\sin x + x \left( \dfrac{d}{dx}\sin x \right) \\[8px] &= (1)\sin x + x \,(\cos x) \\[8px] &= \sin x + x\cos x \quad \cmark
\end{align*} \]
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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 $\dfrac{d}{dx}\tan x = \sec^2 x.$
\[\begin{align*}
\dfrac{d}{dx} \left[\left(e^x +1 \right) \tan x \right] &= \left[\dfrac{d}{dx} \left(e^x +1 \right) \right] \tan x + \left(e^x +1 \right)\left[ \dfrac{d}{dx}\tan x \right] \\[8px] &= \left[ e^x  \right]\tan x + \left(e^x +1 \right)\left[ \sec^2 x\right] \\[8px] &= e^x \tan x + \left(e^x +1 \right)\sec^2 x \quad \cmark
\end{align*} \]
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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 $\dfrac{d}{dx}x = 1.$
\[ \begin{align*}
\dfrac{d}{dx}\left( \dfrac{\sin x}{x}\right) &= \frac{\left( \dfrac{d}{dx}\sin x \right)x – \sin x \left( \dfrac{d}{dx}x \right)}{x^2} \\[8px] &= \frac{(\cos x)x – \sin x \,(1)}{x^2} \\[8px] &= \frac{x \cos x – \sin x}{x^2} \quad \cmark
\end{align*} \]
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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 $\dfrac{d}{dx}(x+1) = 1.$
\[ \begin{align*}
\dfrac{d}{dx} \left(\frac{e^x}{x+1} \right) &= \frac{\left( \dfrac{d}{dx}e^x \right)(x+1) – e^x \left(\dfrac{d}{dx}(x+1) \right)}{(x+1)^2} \\[8px] &= \frac{\left( e^x \right)(x+1) – e^x(1)}{(x+1)^2} \\[8px] &= \frac{e^x(x+1 -1)}{(x+1)^2} \\[8px] &= \frac{xe^x}{(x+1)^2} \quad \cmark
\end{align*} \]
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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
\[ \begin{align*}
\dfrac{d}{dx}\left(\frac{3x}{5 – \tan x} \right) &= \frac{\left(\dfrac{d}{dx}(3x) \right)(5 – \tan x) – (3x)\left(\dfrac{d}{dx}(5 – \tan x) \right)}{(5 – \tan x)^2} \\[8px] &= \frac{(3)(5 – \tan x) – (3x)(-\sec^2 x)}{(5 – \tan x)^2} \\[8px] &= \frac{15 – 3\tan x + 3x\sec^2 x}{(5 – \tan x)^2} \quad \cmark
\end{align*} \]
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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!

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Ibrahim
1 month ago

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

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Matheno
Editor
Reply to  Ibrahim
1 month 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!

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Pratistha
8 months ago

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

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Matheno
Editor
Reply to  Pratistha
8 months ago

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

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Rushi
9 months ago

It’s very much helpful

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Matheno
Editor
Reply to  Rushi
9 months ago

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

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