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+1 vote

Differentiate w.r.t. \(x\) the function in \(\large\frac{\cos^{ -1}\Large\frac{ x}{ 2}}{\sqrt {2x + 7}}, \normalsize-2 < x < 2 \)

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Toolbox:
  • $\big(\large\frac{u}{v}\big)'=\large\frac{u'v-uv'}{v^2}$
Step 1:
y=\(\large\frac{\cos^{ -1}\Large\frac{ x}{ 2}}{\sqrt {2x + 7}}, \normalsize-2 < x < 2 \)
It is of the form $\big(\large\frac{u}{v}\big)$
$\Rightarrow \big(\large\frac{u}{v}\big)'=\large\frac{u'v-uv'}{v^2}$
$u=\cos^{-1}\large\frac{x}{2}$
Differentiating with respect to $x$
$\large\frac{du}{dx}=\frac{1}{\sqrt{1-\Large\frac{x^2}{4}}}$$\times \large\frac{1}{2}=\frac{-1}{\sqrt{4-x^2}}$
Step 2:
$v=\sqrt{2x+7}$
$v'=\large\frac{dv}{dx}=\large\frac{1}{2}$$(2x+7)^{\large\frac{1}{2}-1}\times 2$
$\Rightarrow (2x+7)^{-\large\frac{1}{2}}$
$\Rightarrow \large\frac{1}{\sqrt{2x+7}}$
Step 3:
$\large\frac{dy}{dx}=\big(\large\frac{u}{v}\big)'=\Large\frac{\Large\frac{du}{dx}\normalsize\times\large v-u\times \Large\frac{dv}{dx}}{v^2}$
$\quad\;=\Large\frac{\Large\frac{-1}{\sqrt{4-x^2}}\normalsize\times \sqrt{2x+7}-\Large\frac{\cos^{-1}x/2}{\sqrt{2x+7}}}{(2x+7)}$
$\Rightarrow \Large\frac{\Large\frac{-(2x+7)}{\sqrt{4-x^2}}-\cos^{-1}\Large\frac{x}{2}}{(2x+7)^{\Large\frac{3}{2}}}$
$\quad\;=\Large\frac{-(2x+7)+\sqrt{4-x^2}\cos^{-1}\Large\frac{ x}{ 2}}{(2x+7)^{\Large\frac{3}{2}}\sqrt{4-x^2}}$
answered May 14, 2013 by sreemathi.v
 

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