Exercises 4

This page allows you to practice some exercises on Differentiation. If you notice that you have difficulties, we advise you to go over the corresponding sections in the book.

These questions correspond to the following sections: - Section 6.10 - Exponential Functions - Section 6.11 - Logarithmic Functions

Question 25

Find the first-order derivatives w.r.t. $x$ of:

(a) $y = 5e^x − 3x^3 + 8$

(b) $y = \frac{x}{e^x}$

(c) $y = \frac{x + x^2}{e^x + 1}$

(d) $y = x^3 e^x$

(e) $y = (x + e^x)^2$

(f) $y = \frac{(ax + bx^2)^2}{e^x}$, with $a,b$ constants

Show answer

(a) $y' = 5e^{x} − 9x^{2}$.

(b) $y' = (1 · e^x − x e^x )/e^{2x} = (1 − x)e^{−x}$

(c) $y' = [(1 + 2x)(e^x + 1) − (x + x^2 )e^x ]/(e^x + 1)^2$

$\ \ \ \ \ \ \ \ \ = [1 + 2x + e^x(1 + x − x^2 )]/(e^x + 1)^2$.

(d) $y' = x^{3} e^{x} + 3x^{2} e^{x} = x^2 e^x (3 + x)$.

(e) $y' = 2(x + e^x )(1 + e^x)$.

(f) $y' = [2(ax + bx^2)(a + 2bx)e^{x} − (ax + bx^{2})^2 e^{x}]/(e^{x})^2$

$\ \ \ \ \ \ \ \ \ = [x(a + bx)(−bx^{2} + (4b − a)x + 2a)]e^{−x}$.

Question 26

Find the following derivatives:

(a) $y'$ and $y''$ for $y = e^{1/x}$

(b) $y'$ and $y''$ for $y = 5e^{2x^2 −3x+1}$

(c) $\frac{\text{d}}{\text{d}x} e^{(e^x)}$

(d) $\frac{\text{d}}{\text{d}t} \left( \frac{1}{(e^t + e^{−t})} \right)$

(e) $\frac{\text{d}}{\text{d}x} \ x2^x$

(f) $y'$ for $y = x^2 2^{x^2}$

(g) $y'$ for $y = e^x 10^x$

Show answer

(a) $y' = −x^{−2} e^{1/x}$ and $y'' = x^{−4} e^{1/x} (2x + 1)$.

(b) $y' = 5(4x − 3)e^{2x^2 −3x+1}$ and $y'' = 5e^{2x^{2} −3x+1}(16x^2 − 24x + 13)$.

(c) $\frac{\text{d}}{\text{d}x} e^{(e^x)} = e^{e^{x}} \cdot e^{x} = e^{e^{x} + 1}$.

(d) $\frac{\text{d}}{\text{d}t} \left( \frac{1}{(e^t + e^{−t})} \right) = \frac{1}{2}(e^{t/2} − e^{−t/2})$.

(e) $\frac{\text{d}}{\text{d}x} \ x2^x = 2^{x} + x 2^{x} \ln 2 = 2^x (1 + x \ln 2)$.

(f) $y' = 2x 2^{x^2} (1 + x^{2} \ln 2)$.

(g) $y' = e^{x} 10^{x} + e^{x}10^{x} \ln 10 = e^{x}10^{x} (1 + \ln 10)$.

Question 27

Find the intervals where the following functions are increasing:

(a) $y = x^3 + e^{2x}$

(b) $y = 5x^2 e^{−4x}$

(c) $y = e^x − e^{3x}$

Show answer

(a) $y' = 3x^{2} + 2e^{2x}$. Since $3x^{2} \geq 0$ for all $x$ and $e^{2x} > 0$ for all $x$, $2e^{2x} > 0$ for all $x$, so $y' > 0$ for all $x$, so $y$ is increasing in $(−\infty, \infty)$.

(b) $y' = e^ {−4x} (10x−20x^{2}) = 10x(1−2x)e^{−4x}$. Since $e^{-4x} > 0$ for all $x$, the sign of $y'$ depends on $10x(1-2x)$. The critical points can be found by setting this expression equal to zero, which yields $x=0$ and $x = 1/2$. From this, we can easily deduce that $y$ is increasing in the interval $[0, 1/2]$.

(c) $y' = e^{x}(1 − 3e^{2x})$. Since $e^{x} > 0$ for all $x$, we need to solve for what values of $x$ $1 - 3e^{2x} > 0$. This yields $x < -\frac{1}{2}\ln 3$. Thus, $y$ is increasing in $(−\infty, −\frac{1}{2} \ln 3]$.

Question 28

Compute the first and second derivatives of:

(a) $y = \ln x + 3x − 2$

(b) $y = x^3 \ln x$

(c) $y = \frac{\ln x}{x}$

(d) $y = x^3 (\ln x)^2$

(e) $y = \frac{x^2}{\ln x}$

Show answer

(a) $y' = 1/x + 3$ and $y'' = -1/x^{2}$.

(b) $y' = 3x^{2} \ln x + x^{3} \cdot x^{-1} = 3x^{2} \ln x + x^{2}$, and the second derivative is given by $y'' = 6x \ln x + 3x^{2} \cdot x^{-1} + 2x = x(6 \ln x + 5)$.

(c) $y' = (1 − \ln x)/x^2$ and $y'' = (2 \ln x − 3)/x^3$ after applying the quotient rule.

(d) $y' = 3x^{2}(\ln x)^{2} + 2 \ln x \cdot x^{-1} \cdot x^{3} = x^{2} \ln x (3 \ln x + 2)$ and $y '' = 6x(\ln x)^2 + 10x \ln x + 2x$.

(e) $y' = [\ln x \cdot 2x - x^{2} \cdot x^{-1}]/(\ln x)^{2} = x(2 \ln x − 1)/(\ln x)^2$ and $y'' = \frac{2(\ln x)^{3} - 3 (\ln x)^{2} + 2 \ln x}{(\ln x)^{4}}$. (Getting the second derivative requires some tedious algebra).

Question 29

Find the derivative of:

(a) $y = \ln(\ln x)$

(b) $y = \ln \sqrt{1 − x^2}$

(c) $y = e^{x^3} \ln x^2$

Show answer

(a) $y' = \frac{1}{\ln x} \cdot (1/x) = \frac{1}{x\ln x}$, using the chain rule.

(b) $y = \ln\sqrt{1 - x^{2}} = \frac{1}{2}\ln (1- x^{2})$. So, $y' = \frac{1}{2} (1-x^{2})^{-1} \cdot -2x = -\frac{x}{(1-x^{2})}$, using the chain rule.

(c) We can apply the product and chain rule to obtain: $y' = e^{x^{3}} \cdot \frac{2}{x} + \ln x^{2} \cdot 3x^{2} e^{x^{3}}$. This can be rewritten as $y' = e^{x^{3}} \left(3x^{2} \ln x^{2} + \frac{2}{x}\right) = 2e^{x^{3}} \left(3 x^{2} \ln x + \frac{1}{x} \right)$.

Remark: It is not done in this solution, but note that you can also rewrite $y$ as $y = 2e^{x^{3}}\ln x$ and then proceed to take the first derivative. You could verify that this yields the same solution.

Question 30

Determine the domains of the functions defined by:

(a) $y = \ln(x + 1)$

(b) $y = \ln \left(\frac{3x − 1}{1 − x} \right)$

(c) $y = \ln |x|$

(d) $y = \ln(\ln x)$

(e) $y = \frac{1}{\ln(\ln x) - 1}$

Show answer

Note that $\ln z$ is only defined for $z > 0$. We can use this result to solve the following questions.

(a) $y = \ln(x +1)$, so $x + 1 > 0$ for $x > -1$.

(b) The critical points can be found by solving $3x - 1 = 0$ and $1 - x = 0$, which gives $x= \frac{1}{3}$ and $x = 1$. For those values, the function is not defined. For $x < \frac{1}{3}$, we evaluate the $\ln$ function at a negative value, so the function is not defined. This is also the case for $x > 1$. Thus, the function is only defined for $\frac{1}{3} < x < 1$.

(c) For all $x \neq 0$, we ensure that the function argument is strictly positive.

(d) We need $x > 0$ for the inner $\ln$ function to be defined. However, this is not enough. We also need the input of the outer $\ln$ function to be positive. Note that $\ln x > 0$ for $x > 1$, which is the domain we seek.

(e) From (d) we know that we need $x > 1$ for $\ln(\ln x)$ to be defined. We only have to exclude the case in which the denominator reduces to zero. This happens for $x = e^{e}$, which should thus be excluded. Thus, the domain is $x > 1, x \neq e^{e}$.

Question 31

Find the intervals where the following functions are increasing:

(a) $y = \ln(4 − x^2)$

(b) $y = x^3 \ln x$

(c) $y = \frac{(1 − \ln x)^2}{2x}$

Show answer

(a) $y$ is defined only in $(−2, 2)$, where $y' = −2x/(4 − x^2) > 0$ if and only if $x < 0$. Thus, $y$ is increasing in $(−2, 0]$.

(b) $y$ is defined for $x > 0$, where $y' = x^2 (3 \ln x + 1) > 0$ if and only if $\ln x > −1/3$. Thus, $y$ is increasing in $[e^{−1/3} , \infty)$.

(c) $y$ is defined for $x > 0$. $y' = (1 − \ln x)(\ln x − 3)/2x^2 > 0$ if and only if $1 < \ln x < 3$. Thus, $y$ is increasing in $[e, e^3]$.

Question 32

Differentiate the following functions using logarithmic differentiation:

(a) $y = (2x)^x$

(b) $y = x^{\sqrt{x}}$

(c) $y =(\sqrt{x})^{x}$

Show answer

We provide an elaborate answer for (a). The remaining questions can be solved analogously.

(a) Note that $\ln y = x \ln(2x)$. If we take the derivative w.r.t $x$ on both sides, we obtain

$\frac{1}{y} \frac{\text{d}y}{\text{d}x} = \frac{\text{d}}{\text{d}x} x\ln(2x)$. The derivative on the righthand side reduces to the expression:

$\frac{\text{d}}{\text{d}x} x\ln(2x) = \ln(2x) + x \cdot \frac{1}{2x} \cdot 2 = \ln(2x) + 1$. Now we should note that

$\frac{\text{d}y}{\text{d}x} = \frac{\text{d}}{\text{d}x} y x\ln(2x)$. Since $y = (2x)^{x}$, we obtain $y' = (2x)^x (1 + \ln 2 + \ln x)$ by also noting $\ln(2x) = \ln 2 + \ln x$.

(b) $x^{\sqrt{x} − \frac{1}{2}} \left(\frac{1}{2} \ln x + 1 \right)$.

(c) $\frac{1}{2} (\sqrt{x})^{x} (\ln x + 1)$.