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Question 1. If \(y=\sin^5\left(\frac{x}{5}\right)\), then \(y'=\)
Write \(y=\left[\sin\left(\frac{x}{5}\right)\right]^5\). Using the chain rule, \(y'=5\left[\sin\left(\frac{x}{5}\right)\right]^4\cos\left(\frac{x}{5}\right)\cdot\frac{1}{5}\). The factors \(5\) and \(\frac{1}{5}\) cancel, so \(y'=\sin^4\left(\frac{x}{5}\right)\cos\left(\frac{x}{5}\right)\). Answer: C.
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Question 2. \(\displaystyle \lim_{h\to0}\frac{(x+h)^\pi-x^\pi}{h}\) is
The expression is the definition of the derivative of \(f(x)=x^\pi\): \(\displaystyle f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\). By the power rule, \(\frac{d}{dx}x^\pi=\pi x^{\pi-1}\). Answer: B.
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Question 3. Find \(\frac{dy}{dx}\) for \(y=\sqrt{x}(3x-1)\).
Rewrite \(y=x^{1/2}(3x-1)=3x^{3/2}-x^{1/2}\). Differentiate term by term: \(y'=\frac{9}{2}x^{1/2}-\frac{1}{2}x^{-1/2}\). Using a common denominator, \(y'=\frac{9x-1}{2\sqrt{x}}\). Answer: A.
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Question 4. \(\displaystyle \frac{d}{dx}\ln\left(\frac{2}{2-x}\right)=\)
Use logarithm properties: \(\ln\left(\frac{2}{2-x}\right)=\ln2-\ln(2-x)\). Then \(\frac{d}{dx}\ln2=0\), and \(\frac{d}{dx}[-\ln(2-x)]=-\frac{-1}{2-x}=\frac{1}{2-x}\). Answer: E.
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Question 5. Assume \(f(7)=0\), \(f'(7)=14\), \(g(7)=1\), and \(g'(7)=\frac{1}{7}\). Find \(h'(7)\) given \(h(x)=\frac{f(x)}{g(x)}\).
Apply the quotient rule: \(\displaystyle h'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}\). At \(x=7\), \(\displaystyle h'(7)=\frac{14(1)-0\left(\frac17\right)}{1^2}=14\). Answer: C.
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Question 6. Find the derivative of \(\displaystyle f(x)=\ln\left(\frac{\sqrt{x^2+1}}{x(2x^3-1)^2}\right)\).
Expand with logarithm laws: \(\displaystyle f(x)=\frac12\ln(x^2+1)-\ln x-2\ln(2x^3-1)\). Differentiate: \(\displaystyle f'(x)=\frac12\cdot\frac{2x}{x^2+1}-\frac1x-2\cdot\frac{6x^2}{2x^3-1}\). Therefore \(\displaystyle f'(x)=\frac{x}{x^2+1}-\frac1x-\frac{12x^2}{2x^3-1}\). Answer: D.
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Question 7. A function is defined by the properties: \(f''(x)=18x\), \(f'(1)=5\), and \(f(3)=80\). Find \(f(x)\).
Integrate \(f''(x)=18x\): \(f'(x)=9x^2+C_1\). Use \(f'(1)=5\): \(9+C_1=5\), so \(C_1=-4\). Integrate again: \(f(x)=3x^3-4x+C_2\). Use \(f(3)=80\): \(81-12+C_2=80\), so \(C_2=11\). Thus \(f(x)=3x^3-4x+11\). Answer: E.
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Question 8. \(\displaystyle \frac{d}{dx}\ln\left(\frac{e}{e-x}\right)=\)
Use logarithm properties: \(\ln\left(\frac{e}{e-x}\right)=\ln e-\ln(e-x)=1-\ln(e-x)\). Differentiating gives \(\displaystyle 0-\frac{-1}{e-x}=\frac{1}{e-x}\). Answer: C.
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Question 9. Find the derivative of \(y=\sin(x^2)\).
Let \(u=x^2\). Then \(y=\sin u\). By the chain rule, \(\displaystyle \frac{dy}{dx}=\cos u\cdot\frac{du}{dx}=\cos(x^2)\cdot2x=2x\cos(x^2)\). Answer: A.
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Question 10. If \(\displaystyle y=\frac{(4x-3)^2}{\sqrt{x}}\), then \(\frac{dy}{dx}=\)
Rewrite \(y=(4x-3)^2x^{-1/2}\). Using the product rule, \(\displaystyle y'=8(4x-3)x^{-1/2}-\frac12(4x-3)^2x^{-3/2}\). Factor \(\frac{(4x-3)x^{-3/2}}{2}\): \(\displaystyle y'=\frac{(4x-3)x^{-3/2}}{2}[16x-(4x-3)]\). This becomes \(\displaystyle y'=\frac{3(4x-3)(4x+1)}{2x^{3/2}}\). Answer: D.
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Question 11. If \(f(x)=\sqrt{x+2}\), then which one of the following is true for \(f'(x)\)?
The derivative definition is \(\displaystyle f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\). For \(f(x)=\sqrt{x+2}\), \(f(x+h)=\sqrt{x+h+2}\) and \(f(x)=\sqrt{x+2}\). Therefore \(\displaystyle f'(x)=\lim_{h\to0}\frac{\sqrt{x+h+2}-\sqrt{x+2}}{h}\). Answer: B.
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Question 12. Find \(\displaystyle \frac{d}{dx}(x^4-5x+8)^3\) at \(x=\sqrt{2}\).
Let \(u=x^4-5x+8\). By the chain rule, \(\displaystyle \frac{d}{dx}u^3=3u^2(4x^3-5)\). At \(x=\sqrt2\), \(x^4=4\) and \(x^3=2\sqrt2\), so \(u=12-5\sqrt2\) and \(4x^3-5=8\sqrt2-5\). Thus the derivative is \(\displaystyle 3(12-5\sqrt2)^2(8\sqrt2-5)\). Expanding gives \(6456\sqrt2-8670\). Answer: C.
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Question 13. If \(\displaystyle f(x)=\frac{x-7}{2}-\frac{2}{x-5}\), then \(f'(-1)\) is
Rewrite \(f(x)=\frac12(x-7)-2(x-5)^{-1}\). Differentiate: \(\displaystyle f'(x)=\frac12+2(x-5)^{-2}=\frac12+\frac{2}{(x-5)^2}\). At \(x=-1\), \(\displaystyle f'(-1)=\frac12+\frac{2}{36}=\frac12+\frac1{18}=\frac{5}{9}\). Answer: D.
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Question 14. Find the derivative of \(y=\sqrt[3]{x^2+x}\).
Rewrite \(y=(x^2+x)^{1/3}\). Using the chain rule, \(\displaystyle y'=\frac13(x^2+x)^{-2/3}(2x+1)\). Answer: A.
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Question 15. A function is defined by the following properties: \(f''(x)=12x\), \(f'(x)=1\), \(f(2)=16\). Find \(f(x)\).
The printed condition \(f'(x)=1\) conflicts with \(f''(x)=12x\). To match the supplied answer key, interpret it as \(f'(1)=1\). Integrate \(f''(x)=12x\): \(f'(x)=6x^2+C_1\). Using \(f'(1)=1\), \(6+C_1=1\), so \(C_1=-5\). Integrate again: \(f(x)=2x^3-5x+C_2\). Using \(f(2)=16\), \(16-10+C_2=16\), so \(C_2=10\). Thus \(f(x)=2x^3-5x+10\). Answer: B.
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