• November 24th 2008, 10:34 AM
ea12345
please urgent step by step solution
thanks
• November 24th 2008, 10:43 AM
Peritus
$
\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.

$\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}
$
• November 24th 2008, 11:07 AM
Soroban
hELLO, ea12345!

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

Quote:

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

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

Quote:

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

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

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

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

• November 24th 2008, 12:16 PM
ea12345
Thanks soroban you are great