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Question 1. Assume \(f(3)=0\), \(f'(3)=6\), \(g(3)=1\), and \(g'(3)=\frac{1}{3}\). Find \(h'(3)\) given \(h(x)=\frac{f(x)}{g(x)}\).
Use the quotient rule: \(\displaystyle h'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}\). Substitute the values at \(x=3\): \(\displaystyle h'(3)=\frac{6(1)-0\left(\frac13\right)}{1^2}=6\). Answer: D.
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Question 2. A curve is defined by \(y=\ln(\cos^2 3x)\). Find \(\frac{dy}{dx}\).
Let \(u=\cos^2(3x)\). Then \(\displaystyle y'=\frac{u'}{u}\). Differentiate \(u\) using the chain rule: \(\displaystyle u'=2\cos(3x)\bigl[-\sin(3x)\bigr](3)=-6\cos(3x)\sin(3x)\). Therefore \(\displaystyle y'=\frac{-6\cos(3x)\sin(3x)}{\cos^2(3x)}=-6\tan(3x)\). Answer: D.
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Question 3. A function \(f\) is defined by \(\displaystyle f(x)=\frac{e^x-e^{-x}}{2}\). Find \(f'(x)\).
Differentiate term by term. Since \(\frac{d}{dx}e^x=e^x\) and \(\frac{d}{dx}e^{-x}=-e^{-x}\), \(\displaystyle f'(x)=\frac{e^x-(-e^{-x})}{2}=\frac{e^x+e^{-x}}{2}\). Answer: D.
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Question 4. Find the derivative of \(y=\sin^2 x-\cos^2 x\).
Differentiate each term: \(\displaystyle \frac{d}{dx}(\sin^2x)=2\sin x\cos x\), and \(\displaystyle \frac{d}{dx}(-\cos^2x)=-2\cos x(-\sin x)=2\sin x\cos x\). Thus \(y'=4\sin x\cos x\). Using \(\sin 2x=2\sin x\cos x\), \(y'=2\sin 2x\). Answer: A.
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Question 5. Find \(y'\) given \(y=e^{\cot(x^2)}\).
Use the chain rule. Let \(u=\cot(x^2)\), so \(y=e^u\). Then \(y'=e^u u'\). Also, \(\displaystyle u'=-\csc^2(x^2)\cdot 2x\). Therefore \(\displaystyle y'=-2x\csc^2(x^2)e^{\cot(x^2)}\). Answer: C.
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Question 6. If \(\displaystyle f(x)=\frac{x^3(x-2)}{x+1}\), then \(f'(5)\) is
Let \(N(x)=x^3(x-2)=x^4-2x^3\) and \(D(x)=x+1\). Then \(N'(x)=4x^3-6x^2\) and \(D'(x)=1\). By the quotient rule, \(\displaystyle f'(x)=\frac{N'(x)D(x)-N(x)D'(x)}{D(x)^2}\). At \(x=5\), \(N(5)=375\), \(N'(5)=350\), and \(D(5)=6\). Thus \(\displaystyle f'(5)=\frac{350(6)-375}{36}=\frac{1725}{36}=\frac{575}{12}\). Answer: E.
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Question 7. Given \(f(x)=\arctan(e^{2x})\), find \(f'(x)\).
For \(f(x)=\arctan u\), \(\displaystyle f'(x)=\frac{u'}{1+u^2}\). Here \(u=e^{2x}\), so \(u'=2e^{2x}\) and \(u^2=e^{4x}\). Therefore \(\displaystyle f'(x)=\frac{2e^{2x}}{1+e^{4x}}\). Answer: D.
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Question 8. If \(f(x)=\sqrt[5]{x}\), then \(f'(3)=\)
Rewrite \(f(x)=x^{1/5}\). By the power rule, \(\displaystyle f'(x)=\frac15x^{-4/5}=\frac{1}{5x^{4/5}}\). At \(x=3\), \(\displaystyle f'(3)=\frac{1}{5\cdot3^{4/5}}=\frac{1}{5\sqrt[5]{3^4}}\). Answer: E.
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Question 9. Find the derivative: \(\displaystyle s(t)=\csc\left(\frac{t}{2}\right)\).
Use \(\frac{d}{du}(\csc u)=-\csc u\cot u\). Let \(u=\frac{t}{2}\), so \(\frac{du}{dt}=\frac12\). By the chain rule, \(\displaystyle s'(t)=-\csc\left(\frac t2\right)\cot\left(\frac t2\right)\cdot\frac12\). Thus \(\displaystyle s'(t)=-\frac12\csc\left(\frac t2\right)\cot\left(\frac t2\right)\). Answer: E.
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Question 10. Find \(\frac{dy}{dt}\) when \(t=3\), given \(\displaystyle y=\frac{u^2+1}{u}\) and \(u=\sqrt{t+1}\).
Rewrite \(y=\frac{u^2+1}{u}=u+\frac1u\). Then \(\displaystyle \frac{dy}{du}=1-\frac1{u^2}\). Since \(u=(t+1)^{1/2}\), \(\displaystyle \frac{du}{dt}=\frac{1}{2\sqrt{t+1}}=\frac{1}{2u}\). When \(t=3\), \(u=\sqrt4=2\). Therefore \(\displaystyle \frac{dy}{dt}=\left(1-\frac14\right)\left(\frac14\right)=\frac34\cdot\frac14=\frac{3}{16}\). Answer: A.
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Question 11. A function is defined by the following properties: \(f'(x)=6x\) and \(f(1)=-1\). Find \(f(x)\).
Integrate \(f'(x)=6x\): \(\displaystyle f(x)=3x^2+C\). Use \(f(1)=-1\): \(3(1)^2+C=-1\), so \(C=-4\). Therefore \(f(x)=3x^2-4\). Answer: A.
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Question 12. If \(y=e^{1/x}\), then \(y'=\)
Use the chain rule. Let \(u=\frac1x=x^{-1}\). Then \(y=e^u\), so \(y'=e^u u'\). Since \(u'=-x^{-2}=-\frac1{x^2}\), \(\displaystyle y'=-\frac{e^{1/x}}{x^2}\). Answer: A.
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Question 13. Find \(\frac{dy}{dx}\) if \(y=x2^x\).
Use the product rule on \(y=x\cdot2^x\): \(\displaystyle y'=1\cdot2^x+x\cdot(2^x\ln2)\). Factor out \(2^x\): \(\displaystyle y'=2^x(1+x\ln2)=2^x(x\ln2+1)\). Answer: D.
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Question 14. If \(\displaystyle y=\frac{-4}{\sqrt[3]{x+5}}\), then \(\frac{dy}{dx}=\)
Rewrite \(\displaystyle y=-4(x+5)^{-1/3}\). Using the power and chain rules, \(\displaystyle y'=-4\left(-\frac13\right)(x+5)^{-4/3}\cdot1=\frac43(x+5)^{-4/3}\). Therefore \(\displaystyle y'=\frac{4}{3(x+5)^{4/3}}=\frac{4}{3\sqrt[3]{(x+5)^4}}\). Answer: B.
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Question 15. If \(f(x)=e^x\sin^2x\) for all real numbers \(x\), then \(f'(x)=\)
Use the product rule with \(u=e^x\) and \(v=\sin^2x\). Then \(u'=e^x\) and \(\displaystyle v'=2\sin x\cos x\). Therefore \(\displaystyle f'(x)=e^x\sin^2x+e^x(2\sin x\cos x)\). Factor out \(e^x\): \(\displaystyle f'(x)=e^x(2\sin x\cos x+\sin^2x)\). Answer: B.
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