3.14: Homework- Multirule Derivatives
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Each of the problems below involves a combination of two of the following: product rule, quotient rule, chain rule. Give them a shot!
Answer key note: the answers below are simplified, in some cases more so than I’d expect you to on a quiz or exam, but it is still good practice to try to simplify and see if you got the same thing I did.ans
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\(\frac{d}{dx} \ x^2 \ln(x) e^x\)
\(x^2 \ln(x) e^x + 2x \ln(x) e^x + x e^x\)ans
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\(\frac{d}{dx} \ (2x + 1)^5 (3x - 1)\)
\(3 (2x + 1)^5 + 10(3x-1)(2x + 1)^4\)ans
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\(\frac{d}{dx} \ \sqrt{\sin(x^2)}\)
ans
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\(e^{\frac{1}{x^2 - 1}} \cdot \frac{-2x}{(x^2 - 1)^2}\)ans
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\(\frac{d}{dx} \ \frac{x \ln(x)}{x + 1}\)
\(\frac{\ln(x) + x + 1}{(x+1)^2}\)ans
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\(\frac{d}{dx} \ \cfrac{\cfrac{1}{\ln(x)} + \ln(x)}{\ln(x)}\)
\(\frac{-2}{x (\ln(x))^3}\)ans
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\(\frac{d}{dx} \ e^{\cos(x) \sin(x)}\)
\(e^{\cos(x) \sin(x)} \cdot ((\cos(x))^2 - (\sin(x))^2)\)ans
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\(\frac{d}{dx} \ \sin\left(\frac{x \ln x}{e^x} \right)\)
\(\cos\left( \frac{x \ln x}{e^x} \right) \cdot \frac{\ln(x) + 1 - x \ln(x)}{e^x}\)ans
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\(\frac{d}{dx} \ x^2 \ln(x) e^x\)