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ad-work [718]
3 years ago
9

Evaluate the limit as x goes to infinity of x^3e^-x/5

Mathematics
1 answer:
jekas [21]3 years ago
5 0

Answer: 0

Step-by-step explanation:

\frac{x^{3}e^{-x}}{5}

This could be written as

\frac{x^{3} }{5e^{-x} }

Applying L'H rule and differentiating 3 times , we have

\frac{6}{5e^{x} }

taking the limit as x goes to infinity , we have

0

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Find dy/dx for y= x^3 ln (cot x)
ICE Princess25 [194]
<h3>Answer</h3>

  \dfrac{dy}{dx} = 3x^2 \ln(\cot x)-x^3 \csc(x)\sec(x)

<h3>Explanation</h3>

By the product rule (d/dx)(f(x)g(x)) = f(x)g'(x) + g(x)f'(x), we have

  \begin{aligned}\frac{dy}{dx} &= \left(x^3 \ln (\cot x) \right)' \\&= x^3\big(\ln (\cot x)\big)' + \ln (\cot x) \cdot \left(x^3\right)' \end{aligned}

By the chain rule:

  \begin{aligned}\big(\ln (\cot x)\big)' &= \dfrac{1}{\cot x} \cdot (\cot x)' \\ &= \dfrac{1}{\cot x} \cdot -\csc^2 x\\&= -\tan (x) \csc^2(x) \\&= - \frac{\sin x}{\cos x} \cdot \frac{1}{\sin^2 x} = - \frac{1}{\cos x} \cdot \frac{1}{\sin x} \\&= -\csc(x)\sec(x)\end{aligned}

By the power rule:

  (x^3)' = 3x^2

thus

  \begin{aligned}\frac{dy}{dx} &= x^3\big(\ln (\cot x)\big)' + \ln (\cot x) \cdot \left(x^3\right)' \\&= x^3\big( -\csc(x)\sec(x) \big) + \ln(\cot x) \cdot (3x^2) \\&= -x^3 \csc(x)\sec(x) + 3x^2 \ln(\cot x) \\&= 3x^2 \ln(\cot x)-x^3 \csc(x)\sec(x)\end{aligned}

Nothing to do to simplify any further, other than factoring out x^2.

4 0
3 years ago
What is 95 divided by 5?
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2 years ago
39.2 is what percent p of 112
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39.2 is what percent p of 112

39.2/112

39.2 ÷ 112 = 0.35

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0.35 × 100 = 35
35%

3.92 is 35% of 112.
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The answer is V=384.
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