Computing derivative functions

Exercises · Derivatives · with worked solutions · PDF

Exercise 1

Let \(f:\R\rightarrow\R\), \(f(x)=x^{3}\).

1. Write the difference quotient of \(f\) at the point \(x=1\).

2. Compute \(f'(1)\) as the limit of the difference quotient.

3. Write the equation of the tangent line to the graph of \(f\) at the point \((1,1)\).

Solution

1.

$$
\frac{f(1+h)-f(1)}{h}=\frac{(1+h)^{3}-1}{h}=\frac{3h+3h^{2}+h^{3}}{h}=3+3h+h^{2}
$$

2.

$$
f'(1)=\lim_{h\to0}3+3h+h^{2}=3
$$

1. The tangent line has equation \(y = f(1) + f'(1)(x-1)\), hence

   \[ y=1 + 3(x-1) \]

   from which

   \[ y=3x-2 \]

Exercise 2

Compute the derivative of the following function \(f\), specifying the domain of \(f\) and of \(f'\):

\[ f(x)=x^{2}\sin\left(x^{3}\right) \]

Solution

The functions \(f\) and \(f'\) are defined on all of \({\R}\).

Using the rules for the derivative of a product and of composite functions, we have

\[ f'(x)=2x\cdot\sin\left(x^{3}\right)+x^{2}\cdot\cos\left(x^{3}\right)\cdot3x^{2}=2x\sin\left(x^{3}\right)+3x^{4}\cos\left(x^{3}\right) \]

Exercise 3

Compute the derivative of the following function \(f\), specifying the domain of \(f\) and of \(f'\):

\[ f(x)=x^{3}\cos\left(e^{5x^{2}}\right) \]

Solution

The functions \(f\) and \(f'\) are defined on all of \({\R}\).

Using the rules for the derivative of a product and of composite functions, we have

\[ \begin{array}{l} f'(x)=3x^{2}\cdot\cos\left(e^{5x^{2}}\right)-x^{3}\cdot\sin\left(e^{5x^{2}}\right)\cdot e^{5x^{2}}\cdot10x=\\ \\ = 3x^{2}\cos\left(e^{5x^{2}}\right)-10x^{4}e^{5x^{2}}\sin\left(e^{5x^{2}}\right)\end{array} \]

Exercise 4

Compute the derivative of the following function \(f\), specifying the domain of \(f\) and of \(f'\):

\[ f(x)=e^{1/x} \]

Solution

Using the rules for the derivative of composite functions, we have

\[ f'(x)=-\frac{1}{x^{2}} \cdot e^{1/x} \]

The functions \(f\) and \(f'\) are defined on \({\R}\setminus\{0\}\).

Exercise 5

Compute the derivative of the following function:

\[ f(x) = 3\: x^4 + 5\:x + x^{3/2} - 2 \: x^{-3} \]

Solution

\[ f'(x) = 12\: x^3 + 5 + \frac{3}{2}\sqrt{x} + \frac{6}{x^4} \]

Exercise 6

Compute the derivatives of the following functions:

\[ (1)~~ f(x) = \log \big(|x|\big),~~(2)~~ f(x) =\log (3\:x),~~ (3)~~f(x) =\log \left(\left|\frac{x+2}{3-x}\right|\right) \]

Solution

1.

$$
\left(~\log \big(|x|\big)~\right)'=\frac{1}{|x|} \cdot \sgn (x) = \frac{1}{x}
$$

2.

$$
\left(~ \log (3\:x)~\right)' =\big(\log 3 + \log x \big)' = \frac{1}{x}
$$

3.

\begin{align*}
\left(~\log\left(\left|\frac{x+2}{3-x}\right|\right)~\right)' & = \frac{1}{\left|\frac{x+2}{3-x}\right|} \cdot \sgn\left(\frac{x+2}{3-x}\right) \cdot \left(~\left(\frac{x+2}{3-x}\right)~\right)'= \frac{\left(~\left(\frac{x+2}{3-x}\right)~\right)'}{\frac{x+2}{3-x}} \\[2ex]
&={\left(~\left(\frac{x+2}{3-x}\right)~\right)'} \cdot {\frac{3-x}{x+2}}= \frac{3-x+x+2}{(3-x)^2}\cdot {\frac{3-x}{x+2}}\\[2ex]
&= \frac{5}{(3-x)\cdot(x+2)} = \frac{5}{-x^2+x+6}
\end{align*}

Exercise 7

Compute the derivative of the following function:

\[ f(x) = e^{-3\:x} \: (x^2 + 2\:x -1) \]

Solution

Using the rules for the derivative of a product and of composite functions, we have:

\[ f'(x) = -3\: e^{-3\:x} \: (x^2 + 2\:x -1) + e^{-3\:x} \: (2x + 2) = \]

\[ = e^{-3\:x}(-3x^2-6x+3+2x+2) = e^{-3\:x}(-3x^2-4x+5) \]

Exercise 8

Compute the derivative of the following function:

\[ f(x) = \frac{1}{a} \: \arctan \left( \frac{x}{a} \right),~~~ a>0 \]

Solution

\[ f'(x) \: = \: \frac{1}{a} \cdot \frac{1}{1+x^2/a^2} \cdot \frac{1}{a} \: = \: \frac{1}{a^2}\cdot \frac{1}{\frac{a^2+x^2}{a^2}} \: = \:\frac{1}{a^2+x^2} \]

Exercise 9

Compute the derivative of the following function:

\[ f(x) = x \: \log x \]

Solution

Using the rules for the derivative of a product, we have:

\[ f'(x) = \log x + x \cdot \frac{1}{x} = \log x + 1. \]

Exercise 10

Compute the derivative of the following function:

\[ f(x) = \arctan \left( \frac{1+x}{1-x} \right) \]

Solution

Using the rules for the derivative of a quotient and of composite functions, we have:

\[ f'(x) = \frac{1}{1+\frac{(1+x)^2}{(1-x)^2}} \cdot \frac{(1-x)+(1+x)}{(1-x)^2} \: = \: \frac{(1-x)^2}{(1-x)^2+(1+x)^2} \cdot \frac{2}{(1-x)^2} \: = \]

\[ = \: \frac{2}{1+x^2-2x+1+x^2+2x} = \frac{2}{2(x^2+1)} = \frac{1}{x^2+1} \]

Exercise 11

Compute the derivative of the following function:

\[ f(x) = e^{2\:x} \: (2 \: \sin 3\:x - 4\: \cos 3 \:x) \]

Solution

Using the rules for the derivative of a product and of composite functions, we have

\[ f'(x) = 2\: e^{2\:x} \: (2 \: \sin 3\:x - 4\: \cos 3 \:x) + e^{2\:x} \: (2\cdot3 \: \cos 3\:x + 4\cdot3\: \sin 3 \:x) = \]

\[ = 2\: e^{2\:x} \: (2\: \sin 3\:x - 4\: \cos 3 \:x + 3 \: \cos 3\:x + 6\: \sin 3 \:x) = 2\: e^{2\:x} \: (8 \: \sin 3\:x - \cos 3 \:x) \]

Exercise 12

Compute the derivative of the following function:

\[ f(x) = e^{\frac{x+2}{x-3}} \]

Solution

Using the rules for the derivative of composite functions, we have

\[ f'(x) = e^{\frac{x+2}{x-3}} \left[ \frac{(x-3)-(x+2)}{(x-3)^2}\right] = \frac{-5\: e^{\frac{x+2}{x-3}}}{(x-3)^2} \]

Exercise 13

Compute the derivative of the following function:

\[ f(x) =\tanh x \]

Solution

We compute the derivative of the hyperbolic tangent function

\[ f(x) = \tanh x = \frac{\sinh x}{\cosh x} \]

Using the rules for the derivative of a quotient, we have

\[ f'(x) = \frac{\cosh^2x-\sinh^2x}{\cosh^2 x} = 1 - \tanh^2x \]

Moreover, since \(\cosh^2x-\sinh^2x = 1\), we also have

\[ f'(x) = \frac{\cosh^2x-\sinh^2x}{\cosh^2 x} = \frac{1}{\cosh^2 x} = \sech^2 x, \]

where \(\sech x\) is the hyperbolic secant, defined as \(\sech x = \frac{1}{\cosh x}\).

Exercise 14

Compute the derivative of the following function:

\[ f(x) =\coth x \]

Solution

We compute the derivative of the hyperbolic cotangent function

\[ f(x) = \coth x = \frac{\cosh x}{\sinh x} \]

Using the rules for the derivative of a quotient, we have

\[ f'(x) = \frac{\sinh^2 x-\cosh^2 x}{\sinh^2 x} = 1 - \coth^2 x \]

Moreover, since \(\cosh^2x-\sinh^2x = 1\), we also have

\[ f'(x) = -\frac{\cosh^2x-\sinh^2x}{\sinh^2 x} = -\frac{1}{\sinh^2 x} = - \csch^2 x, \]

where \(\csch x\) is the hyperbolic cosecant, defined as \(\csch x = \frac{1}{\sinh x}\).

Exercise 15

Compute the derivative of the following function:

\[ f(x) = e^{x^2 + 3\:x} \]

Solution

Using the rules for the derivative of composite functions, we have

\[ f'(x) = e^{x^2 + 3\:x}\cdot(2x+3) \]

Exercise 16

Compute the derivative of the following function:

\[ f(x) = \log_2 \big(|3\:x|\big) \]

Solution

For \(x \neq 0\), we have:

\[ f(x)=\left\{\begin{array}{lr} \log_2 \big(3\:x\big), &x>0\\ \\ \log_2 \big(-3\:x\big), & x < 0 \end{array}\right. {\rm ~~~~~~~hence~~~~~~~} f'(x)=\left\{\begin{array}{lr} \frac{3}{3\:x \: \log 2}=\frac{1}{x \: \log 2}, &x>0\\ \\ \frac{-3}{-3\:x \: \log 2}=\frac{1}{x \: \log 2}, & x< 0\end{array}\right. \]

Solution

The function is discontinuous at \(x = 0\), hence it is not differentiable.

Exercise 17

Compute the derivative of the following function:

\[ f(x) = \frac{x^2 + 3\:x -2}{2\:x +1} \]

Solution

Using the rules for the derivative of a quotient, we have

\[\begin{align*} f'(x) &= \: \frac{(2x+3)(2x+1)-2(x^2+3x-2)}{(2x+1)^2} \:= \: \frac{4x^2+2x+6x+3-2x^2-6x+4}{(2x+1)^2}\\[2ex] &=\: \frac{2x^2+2x+7}{(2x+1)^2} \end{align*}\]

Exercise 18

Compute the derivative of the following function:

\[ f(x) = x^2\log(\cos x) \]

Solution

Using the rules for the derivative of composite functions, we have

\[\begin{align*} f'(x) &= \: 2x\cdot\log(\cos x)+x^2\cdot\frac{1}{\cos x}\cdot(-\sin x) = 2x\log(\cos x)-x^2\tan x \end{align*}\]

Exercise 19

Compute the derivative of the following function:

\[ f(x) = \sqrt{\arctan(1+x^2)} \]

Solution

Using the rules for the derivative of composite functions, we have

\[\begin{align*} f'(x) &= \frac{1}{2\sqrt{\arctan(1+x^2)}}\cdot\frac{1}{1+(1+x^2)^2}\cdot2x = \frac{x}{(1+(1+x^2)^2)\sqrt{\arctan(1+x^2)}} \\ &= \frac{x}{(x^4+2x^2+2)\sqrt{\arctan(1+x^2)}} \end{align*}\]

Exercise 20

Determine whether the function

\[ f(x)=e^{-x+1}-2\pi x-1 \]

is invertible and compute the derivative of \(f^{-1}(y)\) for \(y=-2\pi\).

Solution

Since \(f'(x)=-e^{-x+1}-2\pi<0\), for every \(x\in\mathbb{R}\), we immediately obtain that f is strictly monotonically decreasing and hence invertible on \(\mathbb{R}\). We also observe that

\[ e^{-x+1}-2\pi x-1=-2\pi \Longleftrightarrow x=1, \]

i.e., \(f^{-1}(-2\pi)=1\). From the theorem on the derivative of the inverse function we immediately obtain

\[ (f^{-1})'(-2\pi)=\frac{1}{f'(1)}=-\frac{1}{1+2\pi}. \]

Remark: We would not have been able to write the analytic expression of \(f^{-1}\), even though it exists since \(f\) is invertible; the theorem on the derivative of the inverse function allows us to overcome this obstacle, without going through the explicit expression of the inverse.

Exercise 21

Determine whether the function

\[ f(x)=4x+\pi \sin x \]

is invertible and compute the derivative of \(f^{-1}(y)\) for \(y=4\pi\).

Solution

Since \(f'(x)=4+\pi \cos x>0\), for every \(x\in\mathbb{R}\), we immediately obtain that f is strictly monotonically increasing and hence invertible on \(\mathbb{R}\). We also observe that

\[ 4x+\pi \sin x=4\pi \Longleftrightarrow x=\pi, \]

i.e., \(f^{-1}(4\pi)=\pi\). From the theorem on the derivative of the inverse function we immediately obtain

\[ (f^{-1})'(4\pi)=\frac{1}{f'(\pi)}=\frac{1}{4-\pi}. \]

Remark: We would not have been able to write the analytic expression of \(f^{-1}\), even though it exists since \(f\) is invertible; the theorem on the derivative of the inverse function allows us to overcome this obstacle, without going through the explicit expression of the inverse.

Exercise 22

Let \(f \in \mathcal{C}^1(\mathbb{R})\) be such that \(f'(1)=5e\). Setting \(g(x)=f(\log x)\), compute \(g'(e)\).

Solution

We observe that \(g \in \mathcal{C}^1(0,+\infty)\), since it is a composition of functions of class \(\mathcal{C}^1\) on their domains. Therefore, using the theorem on the derivative of composite functions, we obtain

\[ g'(x)=f'(\log x)\frac{1}{x} \quad \Longrightarrow \quad g'(e)=\frac{f'(1)}{e}=5. \]

Exercise 23

Compute the derivative of the following function:

\[ h(x) = x^{x\:\log x} \]

Solution

We use the rule for the derivative of a function raised to another function. The two functions are:

\[ f(x) = x {\rm ~~~~and~~~~} g(x) = x \: \log x \]

Hence:

\[\begin{align*} h'(x) &= \bigg( \exp \left( x \cdot \log^2 x \right) ~\bigg)' = \exp \left( x \cdot \log^2 x \right) \cdot \bigg(x \cdot \log^2 x \bigg)' = \\[2ex] &= x^{x\:\log x} \cdot \left( \log^2 x + x \cdot \bigg( \log^2 x ~\bigg)' \right) = x^{x\:\log x} \cdot \left( \log^2 x + x \cdot \frac{2 \cdot \log x}{x} \right)\\[2ex] &=x^{x\:\log x} \cdot \log x \cdot ( \log x + 2 ) \end{align*}\]

Exercise 24

Compute the derivative of the following function:

\[ f(x) = \frac{a\:x + b}{c\:x + d} \]

Solution

Using the rules for the derivative of a quotient, we have

\[ f'(x) = \frac{a(c\:x+d)-c(a\:x+b)}{(c\:x + d)^2} = \frac{ac\:x + ad - ac\:x - bc}{(c\:x + d)^2} = \frac{ad-bc}{(c\:x + d)^2} \]

Exercise 25

Compute the derivative of the following function:

\[ f(x) = \log \left( |\log x| \right) \]

Solution

For \(x \neq 1\), we have:

\[ f(x)=\left\{\begin{array}{lr} \log \left( \log x \right), &x>1\\ \\ \log \left( -\log x \right), & x < 1 \end{array}\right. {\rm ~~~~~~~hence~~~~~~~} f'(x)=\left\{\begin{array}{lr} \frac{1}{\log x} \cdot \frac{1}{x}=\frac{1}{x \; \log x}, &x>1\\ \\ \frac{1}{-\log x}\cdot - \frac{1}{x}=\frac{1}{x \: \log x}, & x< 1\end{array}\right. \]

The function is discontinuous at \(x = 1\), hence it is not differentiable.

Exercise 26

Compute the derivative of the following function \(f\), specifying the domain of \(f\) and of \(f'\):

\[ f(x)=\sqrt[4]{x^{2}\log(x^{3})} \]

Solution

From \(x^{3}>0\) and \(x^{2}\log(x^{3})\geq0\) it follows that \(f\) is defined for \(x\in[1,+\infty)\), and for such \(x\) we have

\[ f(x)=\sqrt[4]{3x^{2}\log x} \]

By the usual differentiation rules, for \(x>1\) the function \(f\) is differentiable and we have:

\[ f'(x)=\frac{\sqrt[4]{3}}{4\sqrt[4]{x^{6}\log^{3} x}}\left(2x\log x+x\right). \]

For \(x=1\), since the usual differentiation rules cannot be applied (the function \(\sqrt[4]{y}\) is not differentiable at \(y=0\)), we directly examine the limit of the difference quotient, where the increment \(h\) makes sense only if \(h>0\):

\[ \lim_{h\to0^{+}}\frac{f(1+h)-f(1)}{h}=\sqrt[4]{3}\lim_{h\to0^{+}}\frac{\sqrt[4]{(1+h)^{2}\log(1+h)}}{h}= \]

\[ = \sqrt[4]{3}\lim_{h\to0^{+}}\frac{\log^{1/4}(1+h)}{h}=\sqrt[4]{3}\lim_{h\to0^{+}}\frac{h^{1/4}}{h}= \sqrt[4]{3}\lim_{h\to0^{+}}\frac{1}{h^{3/4}}=+\infty \]

where we used \(\log^{1/4}(1+h)\sim h^{1/4}\) for \(h\to0^{+}\). The function \(f\) is not differentiable at \(x=1\). At the point \((1,0)\) the graph has a vertical tangent line.

Alternatively, we reach the same conclusion \(f'(1)=+\infty\) by observing that \(f\) is continuous and that

\[ \lim_{x\to1^{+}}\frac{f(x)-f(1)}{x-1}=\lim_{x\to1^{+}}f'(x)= \lim_{x\to1^{+}}\frac{\sqrt[4]{3}}{4\sqrt[4]{x^{6}\log^{3} x}}\left(2x\log x+x\right)=+\infty \]

using Lagrange's mean value theorem or, equivalently, L'Hopital's rule for the indeterminate forms \(0/0\).

Exercise 27

Give an example of \(f:\R\rightarrow\R\), continuous at \(x=1\), not differentiable at \(x=1\), and such that \(f(0)=2\).

Solution

For example

\[ f(x)=|x-1|+1. \]

The function satisfies \(f(0)=2\) and is continuous on all of \({\R}\).

At \(x=1\) we have \(f'_{+}(1)=1\), \(f'_{-}(1)=-1\), hence the function is not differentiable.