Section 6.4 Derivatives
These answers correspond to the problems in Section 2.2.
(b) \(\ds f'(3)=\lim _{h\to 0}\frac{\frac{1}{2(3+h)-1}-\frac{1}{5}}{h}=\lim _{h\to 0}\frac{-2}{5(5+2h)}=-0.08\text{.}\) (c) \(\ds \lim _{h\to 0}\frac{\sin ^7\left( \frac{\pi }{6}+\frac{h}{2}\right) - \left( \frac{1}{2}\right) ^7}{h}=\left. \frac{d}{dx}\left(\sin^ 7\frac{x}{2}\right)\right| _{x=\frac{\pi }{3}}=\frac{7}{2}\cdot \sin^ 6\frac{\pi }{6} \cdot \cos \frac{\pi }{6}=\frac{7\sqrt{3}}{256}\text{.}\)
\(\ds \lim_{x\to 0^+}\frac{f(x)-f(0)}{x}=1\text{,}\) \(\ds \lim_{x\to 0^-}\frac{f(x)-f(0)}{x}=2\text{.}\)
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 \(x=0\text{.}\)
\(\ds f^\prime(2)=\lim _{x\to 2}\frac{x+\frac{1}{x}-\frac{5}{2}}{x-2}=\frac{3}{4}\text{.}\)
\(2\text{.}\)
\(\ds \frac{1}{2\sqrt{x}}\text{.}\)
Since \(g\) is not differentiable we cannot use the product rule. \(\ds f'(0)=\lim _{h\to 0}\frac{hg(h)}{h}=8\text{.}\)
(b) \(\ds f'(4)=\lim _{h\to 0}\frac{\sqrt{5-(x+h)}-1}{h}=-0.5\text{.}\)
\(\ds F'(0)= \lim _{h\to 0}\frac{\frac{f(h)\sin ^2h}{h}}{h} =\lim _{h\to 0}\frac{f(h)\sin ^2h}{h^2}=f(0)\text{.}\)
\(m=e\text{,}\) \(b=0\text{.}\) 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 \(b\text{.}\)
(a) \(\ds S^\prime(3)=\frac{F'(3)G(3)-F(3)G'(3)}{[G(3)]^2}=-\frac{1}{4}\text{.}\) (b) \(\ds T'(0) =F'(G(0))\cdot G'(0)=0\text{.}\) (c) \(\ds U'(3)=\frac{F'(3)}{F(3)}=-\frac{1}{2}\text{.}\)
(a) \(y=5-x\text{.}\) (b) \(-2\text{.}\) (c) \(\ds -\frac{1}{4}\text{.}\) (d) \(-3\text{.}\) (e) \(-4\text{.}\)
From \(h(1)=f(1)g(1)\) and \(h^\prime (1)=f^\prime (1)g(1)+f(1)g^\prime (1)\) it follows that \(g^\prime(1)=9\text{.}\)
\(2f(g(1))\cdot f'(g(1))\cdot g'(1)=120\text{.}\)
\(\ds f'(x)=-\frac{2}{(x-2)^2}\text{.}\)
\(\ds f^\prime(x)=\sec ^2x\text{.}\) 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 \(\ds g'\left( \frac{\sqrt{\pi }}{2}\right)=2\sqrt{\pi }+\frac{16\sqrt{\pi }}{16+\pi ^2}\text{.}\)
Recall that if \(f(x)=e^x\) then \(f^{-1}(x)=\ln x\text{.}\)
Recall that if \(f(x)=\sin^{-1} x\) is the inverse function of \(g(x)=\sin x\text{,}\) \(x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\text{.}\)
\(f^\prime(1)=g(1)=2\text{.}\)
\(\ds \left. \frac{d}{dx}\left(\sqrt{x}+x^7\right)\right| _{x=1}=\frac{15}{2}\text{.}\)
\(f(x)=x^6\text{,}\) \(a=2\text{.}\)
\(f^\prime(x)=\left(-\sin x+g^\prime(x)\cdot\cos x\right)\cdot e^{g(x)}\text{.}\)
\(\ds f^\prime(x)=(\cos x)\ln (g(x))+\frac{g^\prime(x)\cdot\sin x}{g(x)}\text{.}\)
\(f'(0)=0\text{.}\) 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.
\(f'(x)=2-\sin x>0\) for all \(x\in \mathbb{R}\text{.}\) Let \(g(0)=\alpha\text{.}\) Then \(\ds g'(0)=\frac{1}{f'(g(0))}=\frac{1}{2-\sin \alpha }\text{.}\)
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 \(\ds \frac{d}{dx}(\sin ^{-1}x)=\frac{1}{\sqrt{1-x^2}}\text{.}\)
Use the chain rule and the given property of \(f'(x)\) to get \((1+(f(g(x)))^2)\cdot g'(x)=1\text{.}\)
Write \(\ds y=\frac{1}{2}\cdot (2x^2-2\sqrt{x^4-1})\text{.}\)
0. Note \(\ds f(x)=(x+2)(x^2+4)\text{.}\)
\(\ds y'=-\frac{5!}{x^6}-2^5\sin 2x\text{.}\)
\(A=0\text{,}\) \(B=1\text{.}\)
\(A=0\text{,}\) \(B=1\text{.}\)
Use the chain rule to differentiate \(f^2-g^2\text{.}\)
Use the product rule twice.
\(\ds y'=\left( \frac{3\ln (x+2)}{x}+\frac{3\ln x}{x+2}-\frac{x}{x^2+1}\right) \cdot \frac{(x+2)^{3\ln x}}{(x^2+1)^{1/2}}\text{.}\) Use the logarithmic differentiation.
\(\ds y=\frac{2}{\sqrt{x}}\sinh \sqrt{x} \cdot e^{4\cosh \sqrt{x}}\text{.}\)
From \(\ds f^\prime (x)=\frac{2x+1}{\sqrt{1-(x^2+x)^2}}+5^x\ln 5\) it follows that \(f^\prime (0)=1+\ln 5\text{.}\)
\(e^2\text{.}\)
\(2\text{.}\)
\(\ds \frac{5}{2}\text{.}\)
\(0\text{.}\) Write as a product.
\(\displaystyle \ds \frac{3e^{-3t}\sinh (e^{-3t})}{\sqrt{1- \cosh ^2(e^{-3t})}}\)
\(\ds y^\prime(u)=\frac{1}{3}\left( \frac{1}{u+1}+\frac{1}{u+2}-\frac{2u}{u^2+1}-\frac{2u}{u^2+2}\right)\cdot \left( \frac{(u+1)(u+2)}{(u^2+1)(u^2+2)}\right) ^{1/3}\text{.}\)
\(\ds y^\prime=\frac{x\ln(ex^2)\sinh (\arcsin (x^2\ln x))}{\sqrt{1-x^4\ln ^2x}}\text{.}\)
\(\ds y^\prime=- \frac{10\cdot 7^{1-5x}\ln 7}{\sin \left(2\cdot 7^{1-5x}\right)}\text{.}\)
\(\ds y'=-\frac{4e^{-4x}}{\sqrt{1-e^{8x}}}\text{.}\)
\(\ds y'=-2xe^{\cos x^2}\sin x^2\text{.}\)
\(\ds y'=x^{19}\left( 20\arctan x+ \frac{x}{1+x^2}\right)\text{.}\)
\(\ds y'=2x^{\ln x -1}\ln x\text{.}\)
\(\ds y'=\frac{6e^{3\ln (2x+1)}}{2x+1}\text{.}\)
\(\ds y'=2x^{2x}(\ln x+1)\text{.}\)
\(\ds y'=\frac{e^{2x}}{(x^2+1)^3(1+\sin x)^5}\cdot \left( 2-\frac{6x}{x^2+1}-\frac{5\cos x}{1+\sin x}\right)\text{.}\)
\(\ds y'=\frac{2x+2y^2}{3-4xy}\text{.}\)
\(\ds y'=(x\cosh x+\sinh x )x^{\sinh x}\text{.}\)
\(\ds y'=\frac{xy+y^2-1}{3xy^2+3y^3-x^2-xy-1}\text{.}\)
\(y'=\sec (\sinh x)\tan (\sinh x)\cosh x\text{.}\)
\(\ds y'=\frac{ex^{e-1}-e^x}{e^y-ey^{e-1}}\text{.}\)
\(\ds f'(x)=(6x-3x^2-1)e^{-x}\text{.}\)
\(\ds g'(z)=\frac{z\cos \sqrt{z^2+1} }{\sqrt{z^2+1}}\text{.}\)
\(\ds h'(y)=-\frac{y\tan y+1}{2y}\sqrt{\frac{\cos y}{y}}\text{.}\)
\(\ds f'(x)=\frac{1-x^2}{(x^2+1)^2}\text{.}\)
\(\ds g'(x)=\frac{x}{x^2+1}+4\cot x\text{.}\)
\(\ds f'(x)=\frac{1}{2\sqrt{x}(1+x)}\text{.}\)
\(\ds f'(x)=\frac{5\sinh (5\ln x)}{x}\text{.}\)
\(\ds f'(x)=3\cdot 10^{3x}\cdot \ln 10\text{.}\)
\(f'(x)=x^{9}(10\tanh x+x\mbox{ sech} ^2x)\text{.}\)
\(\ds f'(x)=\left( \frac{\cos x}{x}-\sin x\ln x\right) x^{\cos x}\text{.}\)
\(\ds y^\prime=\frac{(2x^3-1)\sin x-x\cos x}{x^2\sin^2x}\cdot e^{x^2+1}\text{.}\)
\(\ds f^\prime(x)=(2\ln x +1)x^{x^2+1}\text{.}\)
\(\ds f^\prime(x)=-3\tan 3x\text{.}\)
\(\ds f'(x)=\frac{(5-x)(x-1)}{(x+1)^{4}}\text{.}\)
\(f'(x)=2^{2x+1}\ln 2-\frac{4x}{3\sqrt[3]{x^2+1}}\text{.}\)
\(f'(x)=4x\tan (x^2)\cdot \sec ^2x^2\text{.}\)
\(\ds f'(x)=\left( \frac{\ln x}{1+x^2}+\frac{\arctan x}{x}\right) \cdot x^{\arctan x}\text{.}\)
\(f'''(x)=8\cosh (2x)\text{.}\)
\(\ds f'(x)=5+5x^4+5^x\ln 5+\frac{1}{5\sqrt[5]{x^4}}\text{.}\)
\(y'=x^{9}(10\tanh x+x\mbox{ sech} ^2x)\text{.}\)
\(\ds y'=\left( \frac{\cos x}{x\ln x}-\sin x\ln \ln x\right) \cdot (\ln x)^{\cos x}\text{.}\)
\(f'(x)=\coth x\text{.}\)
\(f'(x)=(\cos x-x\sin x)e^{x\cos x}\text{.}\)
\(\ds f'(x)=\frac{1+\cos x}{(1+\cos x)^2}\text{.}\)
\(f'(x)=(\ln x+1)x^x\text{.}\)
\(\ds f'(x)=\frac{3}{x^4}\text{.}\)
\(\ds f'(x)=2x\sin ^2(2x^2)+ 8x^3\sin (2x^2)\cos (x^2)\text{.}\)
\(\ds f'(x)=\left( \ln (x+2)+\frac{x}{x+2}\right) (x+2)^x\text{.}\)
\(\ds y'=\frac{x\sec \sqrt{x^2+1}\tan \sqrt{x^2+1}}{\sqrt{x^2+1}}\text{.}\)
\(\ds y'=\left( e^x\ln x+\frac{e^x}{x}\right) x^{e^x}\text{.}\)
\(\ds y'=3x^2+3^x\ln 3+3(\ln x+1)x^{3x}\text{.}\)
\(\ds y'=-(e^{-2x}+4e^{-8x})\text{.}\)
\(\ds y^\prime=\frac{1}{x\sqrt{x^2-1}}\text{.}\)
\(\ds y'=\left( \frac{5}{x}+3x^2+\frac{2x}{3(x^2+1)}-\frac{4}{x+1}\right) \cdot \frac{x^5e^{x^3}\sqrt[3]{x^2+1}}{(x+1)^4}\text{.}\)
\(\ds f'(x)=\frac{1}{\sec (x^2+7x)}\cdot \left( \frac{2x-3}{x^2-3x+8}-(2x+7)\cdot \ln (x^2-3x+8)\tan (x^2+7x)\right)\text{.}\)
\(\displaystyle \ds f'(x)= \frac{2\sinh (2x-3)}{1+\cosh ^2(2x-3)}\)
\(\ds f'(x)=-3e^{3x-4}\sin (e^{3x-4})\text{.}\)
\(\ds f'(x)=\left( (x^{-1}+2x)\ln \tan x+\frac{\ln x+x^2}{\sin x\cos x}\right) \cdot (\tan x)^{\ln x+x^2}\text{.}\)
\(\ds f'(x)=0\text{.}\)
\(\displaystyle \ds h'(t)=-\frac{1}{3}\sec ^2\left( \frac{t}{3}\right) \cdot e^{-\tan \left( \frac{t}{3}\right) }\)
\(\ds y^\prime = -\frac{3}{8x\ln ^2x}\left( \frac{1}{2\ln x}\right) ^{-1/4}\text{.}\) Note that \(\ds y=\left( \frac{1}{2\ln x}\right) ^{3/4}\text{.}\)
\(\displaystyle \ds f^\prime (y)=\frac{\ln 3}{\ln 7 \cdot \sqrt{1-y^2}\cdot \arcsin y}\cdot 3^{\log _7(\arcsin y)}\)
\(\ds f^\prime (x)=\frac{2x+1}{\sqrt{1-(x^2+x)^2}}+5^x\ln 5\text{.}\)
\(\ds g^\prime (x)=-\frac{3x^2+4x+3}{2\sqrt{x+1}(x^2-3)^2}\sinh \left( \frac{\sqrt{x+1}}{x^2-3}\right)\text{.}\)
\(\ds f^\prime(c)=-3^{\cos x}\cdot(\sin x\cdot \ln 3+2)\cdot e^{-2x}\text{.}\)
\(\ds f^\prime (x)=\frac{2^x(e^{4x}+a)\ln 2 -4e^{4x}\sinh ^{-1}(2^x)\sqrt{2^{2x}+1}}{(e^{4x}+a)^2\sqrt{2^{2x}+1}}\text{.}\)
\(\ds g^\prime (x)=g(x)\cdot \left(-\frac{6x\sin (3x^2)}{2+\cos (3x^2)}+\pi -\frac{3}{2}\right)\text{.}\) (Use logarithmic differentiation.)
\(\ds f^\prime(x)=-5^{\cos x}\cdot\left(\ln 5+\cos x\cdot \csc^2x\right)\text{.}\)
\(\ds y^\prime=x^{\arcsin x}\cdot\left(\frac{\ln x}{\sqrt{1-x^2}}+\frac{\arcsin x}{x}\right)\text{.}\)
\(\ds f^\prime(x)=\frac{(1+x)\cos (x^2)+2x^2\sin (x^2)}{\cos^2(x^2)}\cdot e^x\text{.}\)
\(\displaystyle \ds \frac{d^2y}{dx^2}=-\frac{4x^3}{(1+x^4)^2}\)
\(\displaystyle \ds y^\prime =\frac{1}{2}x^{\sqrt{x}-\frac{1}{2}}\ln (e^2x)\)
\(\ds f^\prime(x)=\frac{(1+\ln x)\sin(2x+3)-2x\ln x\cos(2x+3)}{\sin^2(2x+3)}\text{.}\)
\(\ds f^\prime(x)=-\frac{(x^2+x)\sin x+2x-1}{(x^2+x)^2}\cdot e^{\cos x}\text{.}\)
\(\ds f^\prime(x)=\frac{(x^4+4x+5)^{10}}{\sqrt{x^4-x^2+2}\cdot (x^3+x-6)^3}\cdot\left(\frac{40(x^3+1)}{x^4+4x+5}-\frac{x(2x^2-1)}{x^4-x^2+2}-\frac{3(3x^2+1)}{x^3+x-6}\right)\text{.}\)
\(\ds y^{\prime\prime}=e^{e^x+x}\cdot (e^x+1)\text{.}\)
\(96.5\text{.}\)
\(\ds f^\prime(x)=x^{\sqrt{x}}\left(\frac{\sqrt{x}\ln (e^2x)}{2}+1\right)\text{.}\)
\(\ds g^{(11)}(\theta)=\frac{1}{2^{11}}\cdot \sin\left(\frac{\theta}{2}\right)\text{.}\)
\(\ds f^\prime(x)=\frac{5^{\log_2\pi}e^{\cos x}(1-\cos x)\sin x}{\cos^2x}\text{.}\)
\(\ds f^\prime(x)=\cos^x(x)\cdot(\ln(\cos x)-\tan x)\text{.}\)
\(\ds g^{\prime\prime}(t)=12(4t^2+3)^{-3/2}\text{.}\)
\(\ds f^\prime(x)=-6x^{-7}\text{.}\)
\(\ds f^\prime(x)=2x\sin(2x^2)+4x^3\cos(2x^2)\text{.}\)
\(\ds f^\prime(x)=(x+2)^x\left(\frac{x}{x+2}+\ln(x+2)\right)\text{.}\)
\(\ds y^{\prime\prime}=-8(x+2)^{-3}\text{.}\)
\(\ds f^\prime(x)=((x+6)x^5+10e^x)e^x\text{.}\)
\(\ds f^\prime(x)=-3x^2\sin(\sin(x^3))\cos(x^3)\text{.}\)
\(\ds f^\prime(x)=2x-3^x\ln 3+\frac{2-x}{2(1-x)^{3/2}}\text{.}\)
\(\ds g^\prime(z)=\frac{1}{(2z+1)\ln 10\sqrt{\log|2z+1|}}\text{.}\)
\(\ds y^\prime=\frac{((1+x\tan x)\ln x-1)\sec x}{10\ln^2x}\text{.}\)
\(\ds y^\prime =\left(2\cdot7^{2x}\ln7-\frac{1}{2\sqrt{x}}\right)\cosh(7^{2x}-\sqrt{x})\text{.}\)