Derivatives

Exercises 2.2 Derivatives

Recall that if \(f^\prime(a)\) exists then

\begin{equation*} \displaystyle f^\prime(a)=\lim _{h\to 0}\frac{f(a+h)-f(a)}{h}=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}\text{.} \end{equation*}

1.

Assume that \(f(x)\) is a real-valued function defined for all real numbers \(x\) on an open interval whose centre is a certain real number \(a\text{.}\) What does it mean to say that \(f(x)\) has a derivative \(f'(a)\) at \(x=a\text{,}\) and what is the value of \(f'(a)\text{?}\) (Give the definition of \(f^\prime(a)\text{.}\))
Use the definition of \(f^\prime(a)\) you have just given in part (a) to show that if \(\ds f(x)=\frac{1}{2x-1}\) then \(f'(3)=-0.08\text{.}\)
Find \(\ds \lim _{h\to 0}\frac{\sin ^7\left( \frac{\pi }{6}+\frac{h}{2}\right) - \left( \frac{1}{2}\right) ^7}{h}\text{.}\)

Answer

\(\displaystyle f^\prime(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}=\lim_{x\to a}\frac{f(x)-f(a)}{x-a} \text{.}\)
True.
\(frac{7\sqrt{3}}{256}\text{.}\)

Solution

We state the definition of the derivative:
\begin{equation*} \displaystyle f^\prime(a)=\lim _{h\to 0}\frac{f(a+h)-f(a)}{h}=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}.\text{.} \end{equation*}
We have:

\begin{equation*} \ds f'(3)=\lim _{h\to 0}\frac{\frac{1}{2(3+h)-1}-\frac{1}{5}}{h}=\lim _{h\to 0}\frac{-2}{5(5+2h)}=-0.08\text{.} \end{equation*}
We have:

\begin{equation*} \begin{split} \ds \lim _{h\to 0}\frac{\sin ^7\left( \frac{\pi }{6}+\frac{h}{2}\right) - \left( \frac{1}{2}\right) ^7}{h}\amp=\left. \frac{d}{dx}\left(\sin^ 7\frac{x}{2}\right)\right|_{x=\frac{\pi }{3}}\\ \amp=\frac{7}{2}\cdot \sin^ 6\frac{\pi }{6} \cdot \cos \frac{\pi }{6}\\ \amp=\frac{7\sqrt{3}}{256}. \end{split} \end{equation*}

2.

Explain why the function

\begin{equation*} f(x)=\left\{ \begin{array}{ll} x^2+2x+1,\amp \mbox{if } x\leq 0\\ 1+\sin x,\amp \mbox{if } x> 0 \end{array} \right. \end{equation*}

is continuous but not differentiable on the interval \((-1,1)\text{.}\)

Answer

\(\ds \lim_{x\to 0^+}\frac{f(x)-f(0)}{x}=1\text{,}\) \(\ds \lim_{x\to 0^-}\frac{f(x)-f(0)}{x}=2\text{.}\)

3.

Let \(I\) be a bounded function on \(\mathbb{R}\) and define \(f\) by \(f(x)=x^2I(x)\text{.}\) Show that \(f\) is differentiable at \(x=0\text{.}\)

Solution

Let \(|I(x)|\leq M\) for all \(x\in \mathbb{R}\text{.}\) Then for any \(h\not= 0\text{,}\) \(\ds \left| \frac{h^2I(h)}{h}\right| =|hI(h)|\text{.}\) Use the Squeeze Theorem to conclude that \(f\) is differentiable at \(x=0\text{.}\)

4.

Use the definition of the derivative to find \(f^\prime(2)\) for \(f(x)=x+\frac{1}{x}\text{.}\)

Answer

\(\frac{3}{4} \text{.}\)

Solution

\(\ds f^\prime(2)=\lim _{x\to 2}\frac{x+\frac{1}{x}-\frac{5}{2}}{x-2}=\frac{3}{4}\text{.}\)

5.

Use the definition of the derivative to find \(f^\prime(1)\) for \(f(x)=3x^2-4x+1\text{.}\)

Answer

\(2\text{.}\)

6.

Use the definition of the derivative to find the derivative of \(f(x)=\sqrt{x}\text{.}\) Do not use L'Hopital's rule.

Answer

\(\ds \frac{1}{2\sqrt{x}}\text{.}\)

7.

If \(g\) is continuous (but not differentiable) at \(x=0\text{,}\) \(g(0)=8\text{,}\) and \(f(x)=xg(x)\text{,}\) find \(f^\prime(0)\text{.}\)

Solution

Since \(g\) is not differentiable we cannot use the product rule. \(\ds f'(0)=\lim _{h\to 0}\frac{hg(h)}{h}=8\text{.}\)

8.

Using the definition of the derivative of \(f(x)\) at \(x=4\text{,}\) find the value of \(f^\prime(4)\) if \(f(x)=\sqrt{5-x}\text{.}\)

Answer

\(-0.5\text{.}\)

Solution

\(\ds f'(4)=\lim _{h\to 0}\frac{\sqrt{5-(x+h)}-1}{h}=-0.5\text{.}\)

9.

Let \(f\) be a function that is continuous everywhere and let

\begin{equation*} \ds F(x)=\left\{ \begin{array}{lll} \frac{f(x)\sin ^2x}{x}\amp \mbox{if } \amp x\not= 0,\\ 0\amp \mbox{if } \amp x= 0. \end{array} \right. \end{equation*}

Use the definition of derivatives to evaluate \(F^\prime(0)\text{.}\) Your answer should be in terms of \(f\text{.}\)

Answer

\(F'(0)= f(0). \)

Solution

\(\ds F'(0)= \lim _{h\to 0}\frac{\frac{f(h)\sin ^2h}{h}}{h} =\lim _{h\to 0}\frac{f(h)\sin ^2h}{h^2}=f(0)\text{.}\)

10.

The function

\begin{equation*} f(x)=\left\{ \begin{array}{lll} e^x\amp \mbox{if} \amp x\leq 1\\ mx+b\amp \mbox{if} \amp x>1 \end{array} \right. \end{equation*}

is continuous and differentiable at \(x=1\text{.}\) Find the values for the constants \(m\) and \(b\text{.}\)

Solution

\(m=e\text{,}\) \(b=0\text{.}\) Solve \(\ds \lim _{x\to 1^-}e^x=\lim _{x\to 1^+}(mx+b)\) and \(\ds \lim _{x\to 1^-}\frac{e^x-e}{x-1}=\lim _{x\to 1^+}\frac{mx+b-(m+b)}{x-1}\) for \(m\) and \(b\text{.}\)

11.

Suppose the functions \(F(x)\) and \(G(x)\) satisfy the following properties:

\begin{equation*} \begin{array}{lll} F(3)=2,\amp G(3)=4,\amp G(0)=3\\ F^\prime(3)=-1,\amp G^\prime(3)=0,\amp G^\prime(0)=0 \end{array} \end{equation*}

If \(\ds S(x)=\frac{F(x)}{G(x)}\text{,}\) find \(S^\prime(3)\text{.}\) Simplify your answer.
If \(T(x)=F(G(x))\text{,}\) find \(T^\prime(0)\text{.}\) Simplify your answer.
If \(U(x)=\ln (F(x))\text{,}\) find \(U^\prime(3)\text{.}\) Simplify your answer.

Answer

\(\ds S^\prime(3)=\frac{F'(3)G(3)-F(3)G'(3)}{[G(3)]^2}=-\frac{1}{4}\text{.}\)
\(\ds T'(0) =F'(G(0))\cdot G'(0)=0\text{.}\)
\(\ds U'(3)=\frac{F'(3)}{F(3)}=-\frac{1}{2}\text{.}\)

12.

Suppose the functions \(f(x)\) and \(g(x)\) satisfy the following properties:

\begin{equation*} \begin{array}{lll} f(2)=3,\amp g(2)=4,\amp g(0)=2\\ f^\prime(2)=-1,\amp g^\prime(2)=0,\amp g^\prime(0)=3 \end{array} \end{equation*}

Find an equation of the tangent line to the graph of the function \(f\) at the point \((2, f(2))\text{.}\)
If \(\ds h(x)=2f(x)-3g(x)\text{,}\) find \(h^\prime(2)\text{.}\)
If \(\ds k(x)=\frac{f(x)}{g(x)}\text{,}\) find \(k^\prime(2)\text{.}\)
If \(p(x)=f(g(x))\text{,}\) find \(p^\prime(0)\text{.}\)
If \(r(x)=f(x)\cdot g(x)\text{,}\) find \(r^\prime(2)\text{.}\)

Answer

\(y=5-x\text{.}\)
\(-2\text{.}\)
\(\ds -\frac{1}{4}\text{.}\)
\(-3\text{.}\)
\(-4\text{.}\)

13.

Suppose that \(f(x)\) and \(g(x)\) are differentiable functions and that \(h(x)=f(x)g(x)\text{.}\) You are given the following table of values:

\begin{equation*} \begin{array}{|c|c|} \hline h(1)\amp 24\\ \hline g(1)\amp 6\\ \hline f^\prime(1)\amp -2\\ \hline h^\prime(1)\amp 20\\ \hline \end{array} \end{equation*}

Using the table, find \(g^\prime (1)\text{.}\)

Solution

From \(h(1)=f(1)g(1)\) and \(h^\prime (1)=f^\prime (1)g(1)+f(1)g^\prime (1)\) it follows that \(g^\prime(1)=9\text{.}\)

14.

Given \(F(x)=f^2(g(x))\text{,}\) \(g(1)=2\text{,}\) \(g'(1)=3\text{,}\) \(f(2)=4\text{,}\) and \(f'(2)=5\text{,}\) find \(F'(1)\text{.}\)

Answer

\(2f(g(1))\cdot f'(g(1))\cdot g'(1)=120\text{.}\)

15.

Compute the derivative of \(\ds f(x)=\frac{x}{x-2}\) by

using the limit definition of the derivative;
using the quotient rule.

Answer

\(\ds f'(x)=-\frac{2}{(x-2)^2}\text{.}\)

Solution

We compute:

\begin{equation*} \begin{split} f^\prime(x)\amp=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \\ \amp= \lim_{h\to 0} \left[ \frac{\left(\frac{x+h}{x+h-2}\right) - \left(\frac{x}{x-2}\right)}{h}\right] \\ \amp=\lim_{h\to 0} \left[ \frac{(x+h)(x-2) - x(x+h-2)}{(x+h-2)(x-2)} \cdot \frac{1}{h} \right] \\ \amp = \lim_{h\to 0} \left[ \frac{-2h}{(x+h-2)(x-2)} \cdot \frac{1}{h} \right]\\ \amp = \lim_{h\to 0} \left[\frac{-2}{(x+h-2)(x-2)}\right]\\ \amp= \frac{-2}{(x-2)^2}. \end{split} \end{equation*}
We compute:

\begin{equation*} \begin{split} f^\prime(x)\amp=\frac{\left(\frac{d}{dx} x\right)(x-2) - x \left(\frac{d}{dx} (x-2)\right)}{(x-2)^2} \\ \amp= \frac{(x-2) - x}{(x-2)^2}\\ \amp= \frac{-2}{(x-2)^2}. \end{split} \end{equation*}

16.

Write down the formula for the derivative of \(f(x)=\tan x\text{.}\) State how you could use formulas for derivatives of the sine and cosine functions to derive this formula. (DO NOT do this derivation.)
Use the formula given in part (a) to derive the formula for the derivative of the arctangent function.
Use formulas indicated in parts (a) and (b) to evaluate and simplify the derivative of \(g(x)=\tan (x^2)+\arctan (x^2)\) at \(\ds x=\frac{\sqrt{\pi }}{2}\text{.}\) That is, you want to compute a simplified expression for \(\ds g^\prime\left( \frac{\sqrt{\pi }}{2}\right)\text{.}\)

Answer

\(\ds \sec ^2x\text{.}\)
\(\frac{1}{1+x^2}\text{.}\)
\(2\sqrt{\pi }+\frac{16\sqrt{\pi }}{16+\pi ^2}\text{.}\)

Solution

\(\ds f^\prime(x)=\sec ^2x\text{.}\) This follows from \(\ds \tan x=\frac{\sin x}{\cos x}\) by using the quotient rule.
From \(g(x)=\arctan x\text{,}\) \(x\in \mathbb{R}\text{,}\) and \(\ds f'(g(x))\cdot g'(x)=1\text{,}\) we conclude that \(\ds g'(x)=\cos ^2 (g(x))\text{.}\) Next, suppose that \(x>0\) and consider the right triangle with the hypotenuse of the length 1 and with one angle measured \(g(x)\) radians. Then \(\ds \tan g(x)=\tan (\arctan x)=x=\frac{\sin g(x)}{\cos g(x)}=\sqrt{\frac{1-g'(x)}{g'(x)}}\) which implies that \(\ds x^2=\frac{1-g'(x)}{g'(x)}\text{.}\) Thus \(\ds g'(x)=\frac{1}{1+x^2}\text{.}\)
From \(\ds g^\prime(x)=2x\sec x^2+\frac{2x}{1+x^4}\) it follows that \(\ds g'\left( \frac{\sqrt{\pi }}{2}\right)=2\sqrt{\pi }+\frac{16\sqrt{\pi }}{16+\pi ^2}\text{.}\)

17.

Show that \(\ds \frac{d}{dx}\ln x=\frac{1}{x}\text{.}\)

Hint

Recall that if \(f(x)=e^x\) then \(f^{-1}(x)=\ln x\text{.}\)

18.

Show that \(\ds \frac{d}{dx}\sin^{-1}x=\frac{1}{\sqrt{1-x^2}}\text{.}\)

Hint

Recall that if \(f(x)=\sin^{-1} x\) is the inverse function of \(g(x)=\sin x\text{,}\) \(x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\text{.}\)

19.

If \(g(x)=2x^3+\ln x\) is the derivative of \(f(x)\text{,}\) find

\begin{equation*} \lim _{x\to 0}\frac{f(1+x)-f(1)}{x}\text{.} \end{equation*}

Answer

\(f^\prime(1)=g(1)=2\text{.}\)

20.

Find

\begin{equation*} \lim _{x\to 0}\frac{\sqrt{1+x}+(1+x)^7-2}{x}\text{.} \end{equation*}

Answer

\(\ds \left. \frac{d}{dx}\left(\sqrt{x}+x^7\right)\right| _{x=1}=\frac{15}{2}\text{.}\)

21.

Find a function \(f\) and a number \(a\) such that \(\ds \lim_{h\to0}\frac{(2+h)^6-64}{h}=f^\prime(a)\text{.}\)

Answer

\(f(x)=x^6\text{,}\) \(a=2\text{.}\)

22.

If \(g(x)\) is differentiable and \(\ds f(x)=(\cos x)e^{g(x)}\text{,}\) what is \(f^\prime(x)\text{?}\)

Answer

\(f^\prime(x)=\left(-\sin x+g^\prime(x)\cdot\cos x\right)\cdot e^{g(x)}\text{.}\)

23.

If \(g(x)\) is differentiable and \(\ds f(x)=(\sin x)\ln {g(x)}\text{,}\) what is \(f^\prime(x)\text{?}\)

Answer

\(\ds f^\prime(x)=(\cos x)\ln (g(x))+\frac{g^\prime(x)\cdot\sin x}{g(x)}\text{.}\)

24.

Let \(\ds f(x)=x^2\sin \left( \frac{1}{x}\right)\) if \(x\not= 0\text{,}\) and \(f(0)=0\text{.}\) Find \(f^\prime(0)\) (or say why it doesn't exist.)

Hint

Note that, for \(h\not= 0\text{,}\) \(\ds \left| \frac{h^2\sin \frac{1}{h} }{h}\right|=\left| h\sin \frac{1}{h}\right| \leq |h|\text{.}\) Use the Squeeze Theorem.

Answer

\(f'(0)=0\text{.}\)

25.

Let \(f(x)=2x+\cos x\text{.}\) Say why \(f(x)\) is an increasing function for all \(x\text{.}\) If \(g(x)=f^{-1}(x)\text{,}\) calculate \(g^\prime(0)\text{.}\)

Answer

\(f'(x)=2-\sin x>0\) for all \(x\in \mathbb{R}\text{.}\) Let \(g(0)=\alpha\text{.}\) Then \(\ds g'(0)=\frac{1}{f'(g(0))}=\frac{1}{2-\sin \alpha }\text{.}\)

26.

Show that \(\ds \frac{d}{dx}(\sin ^{-1}x)=\frac{1}{\sqrt{1-x^2}}\text{.}\)

Solution

Let \(f(x)=\sin x\text{,}\) \(\ds x\in \left( -\frac{\pi }{2},\frac{\pi }{2}\right)\) . Then, for \(x\in (-1,1)\text{,}\) \(\ds (f^{-1})'(x)=\frac{1}{\cos (f^{-1}(x))}\text{.}\) Suppose that \(x\in (0,1)\) and let \(\alpha =f^{-1}(x)\text{.}\) Consider the right triangle with the hypothenuse of the length 1 and an angle measured \(\alpha\) radians. The length of the leg opposite to the angle \(\alpha\) equals \(\sin \alpha=x\) which implies \(\ds \frac{d}{dx}(\sin ^{-1}x)=\frac{1}{\sqrt{1-x^2}}\text{.}\)

27.

Suppose that \(f\) is a differentiable function such that \(f(g(x))=x\) and \(f^\prime(x)=1+(f(x))^2\text{.}\) Show that \(\ds g^\prime(x)=\frac{1}{1+x^2}\text{.}\)

Hint

Use the chain rule and the given property of \(f'(x)\) to get \((1+(f(g(x)))^2)\cdot g'(x)=1\text{.}\)

28.

If \(\ds y=\frac{\sqrt{x^2+1}-\sqrt{x^2-1}}{\sqrt{x^2+1}+\sqrt{x^2-1}}\text{,}\) show that \(\ds \frac{dy}{dx}=2x- \frac{2x^3}{\sqrt{x^4-1}}\text{.}\)

Hint

Write \(\ds y=\frac{1}{2}\cdot (2x^2-2\sqrt{x^4-1})\text{.}\)

29.

Let \(f\) be a function differentiable on \(\mathbb{R}\) and such that for all \(x\not=2\text{,}\) \(\ds f(x)=\frac{x^4-16}{x-2}\text{.}\) Find \(f^{(4)}(2)\text{.}\)

Hint

Note \(\ds f(x)=(x+2)(x^2+4)\text{.}\)

Answer

\(f^{(4)}(2)=0\text{.}\)

30.

Given \(\ds y=\frac{1}{x}+\cos 2x\text{,}\) find \(\ds \frac{d^5y}{dx^5}\text{.}\) Simplify your answer.

Answer

\(\ds y'=-\frac{5!}{x^6}-2^5\sin 2x\text{.}\)

31.

Find the values of \(A\) and \(B\) that make

\begin{equation*} f(x)=\left\{ \begin{array}{lll} x^2+1\amp \mbox{ if } \amp x\geq 0\\ A\sin x+B\cos x\amp \mbox{ if } \amp x\lt 0 \end{array} \right. \end{equation*}

differentiable at \(x=0\text{.}\)

Answer

\(A=0\text{,}\) \(B=1\text{.}\)

32.

Find the values of \(A\) and \(B\) that make

\begin{equation*} f(x)=\left\{ \begin{array}{lll} x^2+1\amp \mbox{ if } \amp x\lt 0\\ Ax+B\amp \mbox{ if } \amp x\geq 0 \end{array} \right. \end{equation*}

differentiable at \(x=0\text{.}\)

Answer

\(A=0\text{,}\) \(B=1\text{.}\)

33.

If \(f\) and \(g\) are two functions for which \(f^\prime =g\) and \(g^\prime =f\) for all \(x\text{,}\) then prove that \(f^2-g^2\) must be a constant.

Hint

Use the chain rule to differentiate \(f^2-g^2\text{.}\)

34.

Show that if \(f\) and \(g\) are twice differentiable functions (i.e. both have continuous second derivatives) then \((fg)^{\prime\prime}=f^{\prime\prime}g+2f^\prime g^\prime+fg^{\prime\prime}\text{.}\)

Hint

Use the product rule twice.

35.

Find \(y^\prime\) when \(\ds y=\frac{(x+2)^{3\ln x}}{(x^2+1)^{1/2}}\text{.}\)

Hint

Use the logarithmic differentiation.

Answer

\(\ds y'=\left( \frac{3\ln (x+2)}{x}+\frac{3\ln x}{x+2}-\frac{x}{x^2+1}\right) \cdot \frac{(x+2)^{3\ln x}}{(x^2+1)^{1/2}}\text{.}\)

36.

Find \(y^\prime\) when \(\ds y=e^{4\cosh \sqrt{x}}\text{.}\)

Answer

\(\ds y=\frac{2}{\sqrt{x}}\sinh \sqrt{x} \cdot e^{4\cosh \sqrt{x}}\text{.}\)

37.

Find \(f^\prime (0)\) for the function \(f(x)=\sin ^{-1}(x^2+x)+5^x\text{.}\)

Answer

From \(\ds f^\prime (x)=\frac{2x+1}{\sqrt{1-(x^2+x)^2}}+5^x\ln 5\) it follows that \(f^\prime (0)=1+\ln 5\text{.}\)

38.

Let \(\ds f(x)=\log_a(3x^2-2)\text{.}\) For what value of \(a\) is \(f^\prime(1)=3\text{?}\)

Answer

\(e^2\text{.}\)

39.

Let \(\ds f(x)=e^{a(x^2-1)}\text{.}\) For what value of \(a\) is \(f^\prime(1)=4\text{?}\)

Answer

\(2\text{.}\)

40.

Let \(\ds f(x)=\ln((x^2+1)^a)\text{.}\) For what value of \(a\) is \(f^\prime(2)=2\text{?}\)

Answer

\(\ds \frac{5}{2}\text{.}\)

41.

Given

\begin{equation*} y=\frac{\sqrt{1+2x}\sqrt[4]{1+4x}\sqrt[6]{1+6x}\ldots \sqrt[100]{1+100x}}{\sqrt[3]{1+3x}\sqrt[5]{1+5x}\sqrt[7]{1+7x}\ldots \sqrt[101]{1+101x}}\text{,} \end{equation*}

find \(y'\) at \(x=0\text{.}\)

Hint

Write as a product.

Answer

\(0\text{.}\)

42.

Evaluate \(D_t\cos ^{-1}(\cosh (e^{-3t}))\text{,}\) without simplifying your answer.

Answer

\(\ds \frac{3e^{-3t}\sinh (e^{-3t})}{\sqrt{1- \cosh ^2(e^{-3t})}}\)

43.

Use logarithmic differentiation to find \(y^\prime(u)\) as a function of \(u\) alone, where

\begin{equation*} y(u)=\left( \frac{(u+1)(u+2)}{(u^2+1)(u^2+2)}\right) ^{1/3}\text{,} \end{equation*}

without simplifying your answer.

Answer

\(\ds y^\prime(u)=\frac{1}{3}\left( \frac{1}{u+1}+\frac{1}{u+2}-\frac{2u}{u^2+1}-\frac{2u}{u^2+2}\right)\cdot \left( \frac{(u+1)(u+2)}{(u^2+1)(u^2+2)}\right) ^{1/3}\text{.}\)

44.

Given \(y=\tan (\cos ^{-1}(e^{4x}))\text{,}\) find \(\ds \frac{dy}{dx}\text{.}\) Do not simplify your answer.

Answer

\(\ds y'=-\frac{4e^{-4x}}{\sqrt{1-e^{8x}}}\text{.}\)

Find the derivatives of the following functions:

45.

\(y=\cosh (\arcsin (x^2\ln x))\text{.}\)

Answer

\(\ds y^\prime=\frac{x\ln(ex^2)\sinh (\arcsin (x^2\ln x))}{\sqrt{1-x^4\ln ^2x}}\text{.}\)

46.

\(y=\ln (\tan (7^{1-5x}))\text{.}\)

Answer

\(\ds y^\prime=- \frac{10\cdot 7^{1-5x}\ln 7}{\sin \left(2\cdot 7^{1-5x}\right)}\text{.}\)

Find the derivatives of the following functions:

47.

\(y=e^{\cos x^2}\)

Answer

\(\ds y'=-2xe^{\cos x^2}\sin x^2\text{.}\)

48.

\(y=x^{20}\arctan x\)

Answer

\(\ds y'=x^{19}\left( 20\arctan x+ \frac{x}{1+x^2}\right)\text{.}\)

49.

\(y=x^{\ln x}\)

Answer

\(\ds y'=2x^{\ln x -1}\ln x\text{.}\)

Find the derivatives of the following functions:

50.

\(y=e^{3\ln (2x+1)}\)

Answer

\(\ds y'=\frac{6e^{3\ln (2x+1)}}{2x+1}\text{.}\)

51.

\(y=x^{2x}\)

Answer

\(\ds y'=2x^{2x}(\ln x+1)\text{.}\)

52.

\(\ds y=\frac{e^{2x}}{(x^2+1)^3(1+\sin x)^5}\)

Answer

\(\ds y'=\frac{e^{2x}}{(x^2+1)^3(1+\sin x)^5}\cdot \left( 2-\frac{6x}{x^2+1}-\frac{5\cos x}{1+\sin x}\right)\text{.}\)

53.

\(x^2+2xy^2=3y+4\)

Answer

\(\ds y'=\frac{2x+2y^2}{3-4xy}\text{.}\)

Find the derivatives of the following functions:

54.

\(y=x^{\sinh x}\)

Answer

\(\ds y'=(x\cosh x+\sinh x )x^{\sinh x}\text{.}\)

55.

\(\ln (x+y)=xy-y^3\)

Answer

\(\ds y'=\frac{xy+y^2-1}{3xy^2+3y^3-x^2-xy-1}\text{.}\)

Find the derivatives of the following functions:

56.

\(y=\sec (\sinh x)\)

Answer

\(y'=\sec (\sinh x)\tan (\sinh x)\cosh x\text{.}\)

57.

\(e^x+e^y=x^e+y^e+e^3\)

Answer

\(\ds y'=\frac{ex^{e-1}-e^x}{e^y-ey^{e-1}}\text{.}\)

Find the derivatives of the following functions:

58.

\(\ds f(x)=\frac{3x^2+1}{e^x}\)

Answer

\(\ds f'(x)=(6x-3x^2-1)e^{-x}\text{.}\)

59.

\(g(z)=\sin \sqrt{z^2+1}\)

Answer

\(\ds g'(z)=\frac{z\cos \sqrt{z^2+1} }{\sqrt{z^2+1}}\text{.}\)

60.

\(\ds h(y)=\sqrt{\frac{\cos y}{y}}\)

Answer

\(\ds h'(y)=-\frac{y\tan y+1}{2y}\sqrt{\frac{\cos y}{y}}\text{.}\)

Find the derivatives of the following functions:

61.

\(\ds f(x)=\frac{1}{x+\frac{1}{x}}\)

Answer

\(\ds f'(x)=\frac{1-x^2}{(x^2+1)^2}\text{.}\)

62.

\(g(x)=\ln (\sqrt{x^2+1}\sin ^4x)\)

Answer

\(\ds g'(x)=\frac{x}{x^2+1}+4\cot x\text{.}\)

Find the derivatives of the following functions:

63.

\(f(x)=\arctan (\sqrt{x})\)

Answer

\(\ds f'(x)=\frac{1}{2\sqrt{x}(1+x)}\text{.}\)

64.

\(f(x)=\cosh (5\ln x)\)

Answer

\(\ds f'(x)=\frac{5\sinh (5\ln x)}{x}\text{.}\)

Find the derivatives of the following functions:

65.

\(f(x)=10^{3x}\)

Answer

\(\ds f'(x)=3\cdot 10^{3x}\cdot \ln 10\text{.}\)

66.

\(f(x)=x^{10}\tanh x\)

Answer

\(f'(x)=x^{9}(10\tanh x+x\mbox{ sech} ^2x)\text{.}\)

67.

\(f(x)=x^{\cos x}\)

Answer

\(\ds f'(x)=\left( \frac{\cos x}{x}-\sin x\ln x\right) x^{\cos x}\text{.}\)

Find the derivatives of the following functions:

68.

\(\ds y=\frac{e^{x^2+1}}{x\sin x}\)

Answer

\(\ds y^\prime=\frac{(2x^3-1)\sin x-x\cos x}{x^2\sin^2x}\cdot e^{x^2+1}\text{.}\)

69.

\(f(x)=x^{x^2}\)

Answer

\(\ds f^\prime(x)=(2\ln x +1)x^{x^2+1}\text{.}\)

70.

\(f(x)=\ln (\cos 3x)\)

Answer

\(\ds f^\prime(x)=-3\tan 3x\text{.}\)

Find the derivatives of the following functions:

71.

\(\ds f(x)=\frac{(x-1)^2}{(x+1)^3}\)

Answer

\(\ds f'(x)=\frac{(5-x)(x-1)}{(x+1)^{4}}\text{.}\)

72.

\(f(x)=2^{2x}-(x^2+1)^{2/3}\)

Answer

\(f'(x)=2^{2x+1}\ln 2-\frac{4x}{3\sqrt[3]{x^2+1}}\text{.}\)

73.

\(f(x)=\tan ^2(x^2)\)

Answer

\(f'(x)=4x\tan (x^2)\cdot \sec ^2x^2\text{.}\)

74.

\(f(x)=x^{\arctan x}\)

Answer

\(\ds f'(x)=\left( \frac{\ln x}{1+x^2}+\frac{\arctan x}{x}\right) \cdot x^{\arctan x}\text{.}\)

75.

Compute \(f'''(x)\) where \(f(x)=\sinh (2x)\text{.}\)

Answer

\(f'''(x)=8\cosh (2x)\text{.}\)

Find the derivatives of the following functions:

76.

\(f(x)=5x+x^5+5^x+\sqrt[5]{x}+\ln 5\)

Answer

\(\ds f'(x)=5+5x^4+5^x\ln 5+\frac{1}{5\sqrt[5]{x^4}}\text{.}\)

77.

\(y=x^{10}\tanh x\)

Answer

\(y'=x^{9}(10\tanh x+x\mbox{ sech} ^2x)\text{.}\)

78.

\(y=(\ln x)^{\cos x}\)

Answer

\(\ds y'=\left( \frac{\cos x}{x\ln x}-\sin x\ln \ln x\right) \cdot (\ln x)^{\cos x}\text{.}\)

Find the derivatives of the following functions:

79.

\(f(x)=\ln (\sinh x)\)

Answer

\(f'(x)=\coth x\text{.}\)

80.

\(f(x)=e^{x\cos x}\)

Answer

\(f'(x)=(\cos x-x\sin x)e^{x\cos x}\text{.}\)

81.

\(\ds f(x)=\frac{\sin x}{1+\cos x}\)

Answer

\(\ds f'(x)=\frac{1+\cos x}{(1+\cos x)^2}\text{.}\)

82.

\(f(x)=x^x\)

Answer

\(f'(x)=(\ln x+1)x^x\text{.}\)

Find the derivatives of the following functions:

83.

\(\ds f(x)=g(x^3)\text{,}\) if \(g(x)=\frac{1}{x^2}\)

Answer

\(\ds f'(x)=\frac{3}{x^4}\text{.}\)

84.

\(\ds f(x)=x^2\sin ^2(2x^2)\)

Answer

\(\ds f'(x)=2x\sin ^2(2x^2)+ 8x^3\sin (2x^2)\cos (x^2)\text{.}\)

85.

\(\ds f(x)=(x+2)^x\)

Answer

\(\ds f'(x)=\left( \ln (x+2)+\frac{x}{x+2}\right) (x+2)^x\text{.}\)

Find the derivatives of the following functions:

86.

\(\ds y=\sec \sqrt{x^2+1}\)

Answer

\(\ds y'=\frac{x\sec \sqrt{x^2+1}\tan \sqrt{x^2+1}}{\sqrt{x^2+1}}\text{.}\)

87.

\(\ds y=x^{e^x}\)

Answer

\(\ds y'=\left( e^x\ln x+\frac{e^x}{x}\right) x^{e^x}\text{.}\)

Find the derivatives of the following functions:

88.

\(\ds y=x^3+3^x+x^{3x}\)

Answer

\(\ds y'=3x^2+3^x\ln 3+3(\ln x+1)x^{3x}\text{.}\)

89.

\(\ds y=e^{-5x}\cosh 3x\)

Answer

\(\ds y'=-(e^{-2x}+4e^{-8x})\text{.}\)

90.

\(\ds y=\arctan\left(\sqrt{x^2-1}\right)\)

Answer

\(\ds y^\prime=\frac{1}{x\sqrt{x^2-1}}\text{.}\)

91.

\(\ds y= \frac{x^5e^{x^3}\sqrt[3]{x^2+1}}{(x+1)^4}\)

Answer

\(\ds y'=\left( \frac{5}{x}+3x^2+\frac{2x}{3(x^2+1)}-\frac{4}{x+1}\right) \cdot \frac{x^5e^{x^3}\sqrt[3]{x^2+1}}{(x+1)^4}\text{.}\)

Find the derivatives of the following functions:

92.

\(\ds f(x)=\frac{\ln (x^2-3x+8)}{\sec (x^2+7x)}\)

Answer

\(\ds f'(x)=\frac{1}{\sec (x^2+7x)}\cdot \left( \frac{2x-3}{x^2-3x+8}-(2x+7)\cdot \ln (x^2-3x+8)\tan (x^2+7x)\right)\text{.}\)

93.

\(f(x)= \arctan (\cosh (2x-3))\)

Answer

\(\ds f'(x)= \frac{2\sinh (2x-3)}{1+\cosh ^2(2x-3)}\)

94.

\(\ds f(x)=\cos (e^{3x-4})\)

Answer

\(\ds f'(x)=-3e^{3x-4}\sin (e^{3x-4})\text{.}\)

95.

\(\ds f(x)=(\tan x)^{\ln x+x^2}\)

Answer

\(\ds f'(x)=\left( (x^{-1}+2x)\ln \tan x+\frac{\ln x+x^2}{\sin x\cos x}\right) \cdot (\tan x)^{\ln x+x^2}\text{.}\)

96.

\(\ds f(x)=(\sec ^2x-\tan ^2x)^{45}\)

Answer

\(\ds f'(x)=0\text{.}\)

Find the derivatives of the following functions:

97.

\(\ds h(t)=e^{-\tan \left( \frac{t}{3}\right) }\)

Answer

\(\ds h'(t)=-\frac{1}{3}\sec ^2\left( \frac{t}{3}\right) \cdot e^{-\tan \left( \frac{t}{3}\right) }\)

98.

\(2y^{2/3}= 4y^2\ln x\)

Hint

Note that \(\ds y=\left( \frac{1}{2\ln x}\right) ^{3/4}\text{.}\)

Answer

\(\ds y^\prime = -\frac{3}{8x\ln ^2x}\left( \frac{1}{2\ln x}\right) ^{-1/4}\text{.}\)

99.

\(\ds f(y)=3^{\log _7(\arcsin y)}\)

Answer

\(\ds f^\prime (y)=\frac{\ln 3}{\ln 7 \cdot \sqrt{1-y^2}\cdot \arcsin y}\cdot 3^{\log _7(\arcsin y)}\)

Find the derivatives of the following functions:

100.

\(\ds f(x)=\sin ^{-1}(x^2+x)+5^x\)

Answer

\(\ds f^\prime (x)=\frac{2x+1}{\sqrt{1-(x^2+x)^2}}+5^x\ln 5\text{.}\)

101.

\(g(x)= \cosh \left( \frac{\sqrt{x+1}}{x^2-3}\right)\)

Answer

\(\ds g^\prime (x)=-\frac{3x^2+4x+3}{2\sqrt{x+1}(x^2-3)^2}\sinh \left( \frac{\sqrt{x+1}}{x^2-3}\right)\text{.}\)

102.

\(\ds f(x)=\frac{3^{\cos x}}{e^{2x}}\)

Answer

\(\ds f^\prime(c)=-3^{\cos x}\cdot(\sin x\cdot \ln 3+2)\cdot e^{-2x}\text{.}\)

Find the derivatives of the following functions:

103.

\(\ds f(x)=\frac{\sinh ^{-1}(2^x)}{e^{4x}+a}, \ a\in \mathbb{R}\)

Answer

\(\ds f^\prime (x)=\frac{2^x(e^{4x}+a)\ln 2 -4e^{4x}\sinh ^{-1}(2^x)\sqrt{2^{2x}+1}}{(e^{4x}+a)^2\sqrt{2^{2x}+1}}\text{.}\)

104.

\(\ds g(x)= \frac{(2+\cos (3x^2))e^{\pi x}}{3\sqrt{x}}\)

Hint

Use logarithmic differentiation.

Answer

\(\ds g^\prime (x)=g(x)\cdot \left(-\frac{6x\sin (3x^2)}{2+\cos (3x^2)}+\pi -\frac{3}{2}\right)\text{.}\)

105.

\(\ds f(x)=\frac{5^{\cos x}}{\sin x}\)

Answer

\(\ds f^\prime(x)=-5^{\cos x}\cdot\left(\ln 5+\cos x\cdot \csc^2x\right)\text{.}\)

106.

\(\ds y=x^{\arcsin x}\)

Answer

\(\ds y^\prime=x^{\arcsin x}\cdot\left(\frac{\ln x}{\sqrt{1-x^2}}+\frac{\arcsin x}{x}\right)\text{.}\)

Find the derivatives of the following functions:

107.

\(\ds f(x)=\frac{xe^x}{\cos(x^2)}\)

Answer

\(\ds f^\prime(x)=\frac{(1+x)\cos (x^2)+2x^2\sin (x^2)}{\cos^2(x^2)}\cdot e^x\text{.}\)

108.

Find \(\ds \frac{d^2y}{dx^2}\) if \(y=\arctan (x^2)\text{.}\)

Answer

\(\ds \frac{d^2y}{dx^2}=-\frac{4x^3}{(1+x^4)^2}\)

109.

\(\ds y=x^{\sqrt{x}}\)

Answer

\(\ds y^\prime =\frac{1}{2}x^{\sqrt{x}-\frac{1}{2}}\ln (e^2x)\)

Find the derivatives of the following functions:

110.

\(\ds f(x)=\frac{x\ln x}{\sin(2x+3)}\)

Answer

\(\ds f^\prime(x)=\frac{(1+\ln x)\sin(2x+3)-2x\ln x\cos(2x+3)}{\sin^2(2x+3)}\text{.}\)

111.

\(\ds f(x)=\frac{e^{\cos x}}{x^2+x}\)

Answer

\(\ds f^\prime(x)=-\frac{(x^2+x)\sin x+2x-1}{(x^2+x)^2}\cdot e^{\cos x}\text{.}\)

112.

\(\ds f(x)=\frac{(x^4+4x+5)^{10}}{\sqrt{x^4-x^2+2}}\cdot \frac{1}{(x^3+x-6)^3}\)

Answer

\(\ds f^\prime(x)=\frac{(x^4+4x+5)^{10}}{\sqrt{x^4-x^2+2}\cdot (x^3+x-6)^3}\cdot\left(\frac{40(x^3+1)}{x^4+4x+5}-\frac{x(2x^2-1)}{x^4-x^2+2}-\frac{3(3x^2+1)}{x^3+x-6}\right)\text{.}\)