Section 2.2 Derivatives
Recall that if \(f^\prime(a)\) exists then
Solve the following problems:
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 \(\ds \lim _{h\to 0}\frac{\sin ^7\left( \frac{\pi }{6}+\frac{h}{2}\right) - \left( \frac{1}{2}\right) ^7}{h}\text{.}\)
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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 \((-1,1)\text{.}\)
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 \(x=0\text{.}\)
Use the definition of the derivative to find \(f^\prime(2)\) for \(f(x)=x+\frac{1}{x}\text{.}\)
Use the definition of the derivative to find \(f^\prime(1)\) for \(f(x)=3x^2-4x+1\text{.}\)
Use the definition of the derivative to find the derivative of \(f(x)=\sqrt{x}\text{.}\) Do not use L'Hopital's rule.
If \(g\) is continuous (but not differentiable) at \(x=0\text{,}\) \(g(0)=8\text{,}\) and \(f(x)=xg(x)\text{,}\) find \(f^\prime(0)\text{.}\)
State the definition of the derivative of \(f(x)\) at \(x=a\text{.}\)
Using the definition of the derivative of \(f(x)\) at \(x=4\text{,}\) find the value of \(f^\prime(4)\) if \(f(x)=\sqrt{5-x}\text{.}\)
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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 \(f\text{.}\)
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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 \(b\text{.}\)
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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.
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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 \(r^\prime(2)\text{.}\)
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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 \(g^\prime (1)\text{.}\)
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 \(F'(1)\text{.}\)
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Compute the derivative of \(\ds f(x)=\frac{x}{x-2}\) by
using the limit definition of the derivative;
using the quotient rule.
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 \(\ds g^\prime\left( \frac{\sqrt{\pi }}{2}\right)\text{.}\)
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Show that
\(\ds \frac{d}{dx}\ln x=\frac{1}{x}\text{.}\)
\(\ds \frac{d}{dx}\sin^{-1}x=\frac{1}{\sqrt{1-x^2}}\text{.}\)
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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*} -
Find
\begin{equation*} \lim _{x\to 0}\frac{\sqrt{1+x}+(1+x)^7-2}{x}\text{.} \end{equation*} Find a function \(f\) and a number \(a\) such that \(\ds \lim_{h\to0}\frac{(2+h)^6-64}{h}=f^\prime(a)\text{.}\)
If \(g(x)\) is differentiable and \(\ds f(x)=(\cos x)e^{g(x)}\text{,}\) what is \(f^\prime(x)\text{?}\)
If \(g(x)\) is differentiable and \(\ds f(x)=(\sin x)\ln {g(x)}\text{,}\) what is \(f^\prime(x)\text{?}\)
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.)
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 \(g^\prime(0)\text{.}\)
Show that \(\ds \frac{d}{dx}(\sin ^{-1}x)=\frac{1}{\sqrt{1-x^2}}\text{.}\)
Suppose that \(f\) is a differentiable function such that \(f(g(x))=x\) and \(f^\prime(x)=1+(f(x))^2\text{.}\) Show that \(\ds g^\prime(x)=\frac{1}{1+x^2}\text{.}\)
If \(\ds y=\frac{\sqrt{x^2+1}-\sqrt{x^2-1}}{\sqrt{x^2+1}+\sqrt{x^2-1}}\text{,}\) show that \(\ds \frac{dy}{dx}=2x- \frac{2x^3}{\sqrt{x^4-1}}\text{.}\)
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 \(f^{(4)}(2)\text{.}\)
Given \(\ds y=\frac{1}{x}+\cos 2x\text{,}\) find \(\ds \frac{d^5y}{dx^5}\text{.}\) Simplify your answer.
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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 \(x=0\text{.}\)
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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 \(x=0\text{.}\)
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.
Show that if \(f\) and \(g\) are twice differentiable functions (i.e. both have continuous second derivatives) then \((fg)^{\prime\prime}=f^{\prime\prime}g+2f^\prime g^\prime+fg^{\prime\prime}\text{.}\)
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Find \(y^\prime\) when
\(\ds y=\frac{(x+2)^{3\ln x}}{(x^2+1)^{1/2}}\text{,}\)
\(\ds y=e^{4\cosh \sqrt{x}}\text{.}\)
Find \(f^\prime (0)\) for the function \(f(x)=\sin ^{-1}(x^2+x)+5^x\text{.}\)
Let \(\ds f(x)=\log_a(3x^2-2)\text{.}\) For what value of \(a\) is \(f^\prime(1)=3\text{?}\)
Let \(\ds f(x)=e^{a(x^2-1)}\text{.}\) For what value of \(a\) is \(f^\prime(1)=4\text{?}\)
Let \(\ds f(x)=\ln((x^2+1)^a)\text{.}\) For what value of \(a\) is \(f^\prime(2)=2\text{?}\)
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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 \(x=0\text{.}\)
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The following questions involve derivatives.
Evaluate \(D_t\cos ^{-1}(\cosh (e^{-3t}))\text{,}\) without simplifying your answer.
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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.
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Differentiate
\(y=\cosh (\arcsin (x^2\ln x))\text{,}\)
\(y=\ln (\tan (7^{1-5x}))\text{.}\)
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Given \(y=\tan (\cos ^{-1}(e^{4x}))\text{,}\) find \(\ds \frac{dy}{dx}\text{.}\) Do not simplify your answer.
Find the derivatives of the following functions:
\(\displaystyle y=e^{\cos x^2}\)
\(\displaystyle y=x^{20}\arctan x\)
\(\displaystyle y=x^{\ln x}\)
\(\displaystyle y=e^{3\ln (2x+1)}\)
\(\displaystyle y=x^{2x}\)
\(\displaystyle \ds y=\frac{e^{2x}}{(x^2+1)^3(1+\sin x)^5}\)
\(\displaystyle x^2+2xy^2=3y+4\)
\(\displaystyle y=x^{\sinh x}\)
\(\displaystyle \ln (x+y)=xy-y^3\)
\(\displaystyle y=\sec (\sinh x)\)
\(\displaystyle e^x+e^y=x^e+y^e+e^3\)
\(\displaystyle \ds f(x)=\frac{3x^2+1}{e^x}\)
\(\displaystyle g(z)=\sin \sqrt{z^2+1}\)
\(\displaystyle \ds h(y)=\sqrt{\frac{\cos y}{y}}\)
\(\displaystyle \ds f(x)=\frac{1}{x+\frac{1}{x}}\)
\(\displaystyle g(x)=\ln (\sqrt{x^2+1}\sin ^4x)\)
\(\displaystyle f(x)=\arctan (\sqrt{x})\)
\(\displaystyle f(x)=\cosh (5\ln x)\)
\(\displaystyle f(x)=10^{3x}\)
\(\displaystyle f(x)=x^{10}\tanh x\)
\(\displaystyle f(x)=x^{\cos x}\)
\(\displaystyle \ds y=\frac{e^{x^2+1}}{x\sin x}\)
\(\displaystyle f(x)=x^{x^2}\)
\(\displaystyle f(x)=\ln (\cos 3x)\)
\(\displaystyle \ds f(x)=\frac{(x-1)^2}{(x+1)^3}\)
\(\displaystyle f(x)=2^{2x}-(x^2+1)^{2/3}\)
\(\displaystyle f(x)=\tan ^2(x^2)\)
\(\displaystyle f(x)=x^{\arctan x}\)
Compute \(f'''(x)\) where \(f(x)=\sinh (2x)\text{.}\)
\(\displaystyle f(x)=5x+x^5+5^x+\sqrt[5]{x}+\ln 5\)
\(\displaystyle y=x^{10}\tanh x\)
\(\displaystyle y=(\ln x)^{\cos x}\)
\(\displaystyle f(x)=\ln (\sinh x)\)
\(\displaystyle f(x)=e^{x\cos x}\)
\(\displaystyle \ds f(x)=\frac{\sin x}{1+\cos x}\)
\(\displaystyle f(x)=x^x\)
\(\ds f(x)=g(x^3)\text{,}\) if \(g(x)=\frac{1}{x^2}\)
\(\displaystyle \ds f(x)=x^2\sin ^2(2x^2)\)
\(\displaystyle \ds f(x)=(x+2)^x\)
\(\displaystyle \ds y=\sec \sqrt{x^2+1}\)
\(\displaystyle \ds y=x^{e^x}\)
\(\displaystyle \ds y=x^3+3^x+x^{3x}\)
\(\displaystyle \ds y=e^{-5x}\cosh 3x\)
\(\displaystyle \ds y=\arctan\left(\sqrt{x^2-1}\right)\)
\(\displaystyle \ds y= \frac{x^5e^{x^3}\sqrt[3]{x^2+1}}{(x+1)^4}\)
\(\displaystyle \ds f(x)=\frac{\ln (x^2-3x+8)}{\sec (x^2+7x)}\)
\(\displaystyle f(x)= \arctan (\cosh (2x-3))\)
\(\displaystyle \ds f(x)=\cos (e^{3x-4})\)
\(\displaystyle \ds f(x)=(\tan x)^{\ln x+x^2}\)
\(\displaystyle \ds f(x)=(\sec ^2x-\tan ^2x)^{45}\)
\(\displaystyle \ds h(t)=e^{-\tan \left( \frac{t}{3}\right) }\)
\(\displaystyle 2y^{2/3}= 4y^2\ln x\)
\(\displaystyle \ds f(y)=3^{\log _7(\arcsin y)}\)
\(\displaystyle \ds f(x)=\sin ^{-1}(x^2+x)+5^x\)
\(\displaystyle g(x)= \cosh \left( \frac{\sqrt{x+1}}{x^2-3}\right)\)
\(\displaystyle \ds f(x)=\frac{3^{\cos x}}{e^{2x}}\)
\(\displaystyle \ds f(x)=\frac{\sinh ^{-1}(2^x)}{e^{4x}+a}, \ a\in \mathbb{R}\)
\(\displaystyle \ds g(x)= \frac{(2+\cos (3x^2))e^{\pi x}}{3\sqrt{x}}\)
\(\displaystyle \ds f(x)=\frac{5^{\cos x}}{\sin x}\)
\(\displaystyle \ds y=x^{\arcsin x}\)
\(\displaystyle \ds f(x)=\frac{xe^x}{\cos(x^2)}\)
Find \(\ds \frac{d^2y}{dx^2}\) if \(y=\arctan (x^2)\text{.}\)
\(\displaystyle \ds y=x^{\sqrt{x}}\)
\(\displaystyle \ds f(x)=\frac{x\ln x}{\sin(2x+3)}\)
\(\displaystyle \ds f(x)=\frac{e^{\cos x}}{x^2+x}\)
\(\displaystyle \ds f(x)=\frac{(x^4+4x+5)^{10}}{\sqrt{x^4-x^2+2}}\cdot \frac{1}{(x^3+x-6)^3}\)
Find \(y^{\prime\prime}\) if \(y=e^{e^x}\text{.}\)
\(\ds f(x)=3x^4+\ln x\text{,}\) find \(f^\prime(2)\)
\(\displaystyle f(x)=x^{\sqrt{x}+1}\)
\(g(\theta )=\cos\left(\frac{\theta}{2}\right)\text{,}\) find \(g^{(11)}(\theta)\)
\(\displaystyle \ds f(x)=\frac{5^{\log_2(\pi)}e^{\cos(x)}}{\cos(x)}\)
\(\displaystyle \ds y=\cos^x(x)\)
\(g(t)=\sqrt{4t^2+3}\text{,}\) find \(g^{\prime\prime}(t)\)
\(\ds f(x)=g\left(x^3\right)\text{,}\) if \(\ds g(x)=\frac{1}{x^2}\)
\(\displaystyle \ds f(x)=x^2\sin^2(2x^2)\)
\(\displaystyle \ds f(x)=(x+2)^x\)
\(\ds y=\frac{-4}{x+2}\text{,}\) find \(y^{\prime\prime}\)
\(\displaystyle \ds f(x)=x^6e^x+5e^{2x}\)
\(\displaystyle \ds f(x)=\cos(\sin(x^3))\)
\(\displaystyle \ds f(x)=1-3^x+x^2+\frac{x}{\sqrt{1-x}}\)
\(\displaystyle \ds g(z)=\sqrt{\log(|2z+1|)}\)
\(\displaystyle \ds y=\frac{x\sec x}{5\ln (x^2)}\)
\(\displaystyle \ds y=\sinh (7^{2x}-\sqrt{x})\)