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  1. #1
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    most urgent please

    please urgent step by step solution
    thanks
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  2. #2
    Senior Member Peritus's Avatar
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    $\displaystyle
    \begin{gathered}
    g(x) = \sqrt x f(x) \hfill \\
    \Rightarrow g'(x) = \frac{1}
    {{2\sqrt x }}f(x) + \sqrt x f'(x) \hfill \\
    \end{gathered}
    $

    now use the given data.


    $\displaystyle \begin{gathered}
    \frac{{d^2 }}
    {{dx^2 }}\left( {\frac{1}
    {{\sin x}}} \right) = - \frac{d}
    {{dx}}\sin ^{ - 2} x\cos x = - 2\sin ^{ - 3} x\cos x\sin x = \hfill \\
    = - 2\sin ^{ - 2} x\cos x \hfill \\
    \end{gathered}
    $
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  3. #3
    Super Member

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    hELLO, ea12345!

    These are straight-forward . . . exactly where is your difficulty?


    $\displaystyle \text{Given: }\:y \:=\:\csc x.\qquad\text{ Find: }\:\frac{d^2y}{dx^2}$
    . . $\displaystyle \frac{dy}{dx} \;=\;\underbrace{-\csc x\cdot\cot x}_{\text{product}}$

    $\displaystyle \frac{d^2y}{dx^2} \;=\;(-\csc x)(-\csc^2\!x) + (+\csc x\cot x)(\cot x) \;=\;\csc^3\!x + \csc x\cot^2\!x $




    $\displaystyle \text{If }\,f(4) = 3\,\text{ and }\,f'(4) = \text{-}5,\;\text{ find }g'(4),\;\text{ where }\:g(x) \:=\:\sqrt{x}\!\cdot\!f(x)$
    We have: .$\displaystyle g(x) \;=\;x^{\frac{1}{2}}\cdot f(x) $ . . . a product

    Then: .$\displaystyle g'(x) \;=\;x^{\frac{1}{2}}\!\cdot\!f'(x) + \tfrac{1}{2}x^{\text{-}\frac{1}{2}}\!\cdot\!f(x) \;=\;\sqrt{x}\!\cdot\!f'(x) + \frac{1}{2\sqrt{x}}\!\cdot\!f(x) $


    Hence: .$\displaystyle g'(4) \;=\;\sqrt{4}\!\cdot\!f'(4) + \frac{1}{2\sqrt{4}}\!\cdot\!f(4) $

    . . . . . . . . . $\displaystyle = \quad 2\!\cdot(-5) + \tfrac{1}{4}\cdot(3) \;=\;-10 + \tfrac{3}{4} \;=\;-\frac{37}{4}$

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  4. #4
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    Thanks soroban you are great
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