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Find the second derivative of the function.\newlinef(x)=5x+3x3f(x) = \frac{5}{x} + \frac{3}{x^3}\newlinef′′(x)=f''(x) = ______

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Q. Find the second derivative of the function.\newlinef(x)=5x+3x3f(x) = \frac{5}{x} + \frac{3}{x^3}\newlinef′′(x)=f''(x) = ______
  1. Find Derivative of f(x)f(x): Find the first derivative of f(x)f(x).f′(x)=−5x2−9x4f'(x) = -\frac{5}{x^2} - \frac{9}{x^4}
  2. Find Second Derivative of f(x)f(x): Now, find the second derivative of f(x)f(x).f′′(x)=10x3+36x5f''(x) = \frac{10}{x^3} + \frac{36}{x^5}

More problems from Find second derivatives of trigonometric, exponential, and logarithmic functions

Question
f′(x)=12ex and f(4)=−16+12e4.f(0)= \begin{array}{l}f^{\prime}(x)=12 e^{x} \text { and } f(4)=-16+12 e^{4} . \\ f(0)=\end{array}
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Solve the equation. b6=3b=□\begin{align*} \frac{b}{6} &= 3 \\ b &= \Box \end{align*}
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Solve the equation. \begin{align*} \frac{b}{6} &= 3, \ b &= \Box \end{align*}
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Solve for nn.125=n8\frac{12}{5}=\frac{n}{8} \newline n=□n = \Box
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The second derivative of the function f f is defined by f′′(x)=x2−5cos⁡(2x) f^{\prime \prime}(x)=x^{2}-5 \cos (2 x) for −2.5<x<3.5 -2.5<x<3.5 . Find the x x -values, if any, in the given domain where the function f f has an inflection point. You may use a calculator and round all values to 33 decimal places.\newlineAnswer: x= x=
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f′(x)=12ex and f(4)=−16+12e4.f(0)= \begin{array}{l}f^{\prime}(x)=12 e^{x} \text { and } f(4)=-16+12 e^{4} . \\ f(0)=\end{array}
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x2−6x+11=yx=y+1 \begin{array}{c}x^{2}-6 x+11=y \\ x=y+1\end{array}
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h(x)=x2+xg(x)=3x+5 Find (h⋅g)(x)\begin{array}{l}h(x)=x^{2}+x \\ g(x)=3 x+5 \\ \text{ Find }(h \cdot g)(x)\end{array}
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Factor completely:\newline8c3+27m6 8 c^{3}+27 m^{6} \newlineAnswer:
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