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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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If \( y= [ x + \sqrt{x^2+a^2}]^n. \) then prove that : $\large\frac{dy}{dx}=\frac{ny}{\sqrt{x^2+a^2}}$

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Toolbox:
  • $\large\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}$
Step 1:
$y=[x+\sqrt{x^2+a^2}]^n$
Let $u=x+\sqrt{x^2+a^2}$
Differentiating w.r.t $x$ we get,
$\large\frac{du}{dx}$$=1+\large\frac{1}{2}$$(x^2+a^2)^{\Large\frac{-1}{2}}(2x)$
$\quad\;\;=1+\large\frac{x}{\sqrt{x^2+a^2}}$
Step 2:
$y=u^n$
Differentiating w.r.t $x$ we get,
$\large\frac{dy}{du}$$=nu^{n-1}$
$\large\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}$
$\large\frac{dy}{dx}$$=nu^{n-1}\times (1+\large\frac{x}{\sqrt{x^2+a^2}})$
Substituting for $u$ we get,
$\large\frac{dy}{dx}$$=n[x+\sqrt{x^2+a^2}]^{n-1}\times [1+\large\frac{x}{\sqrt{x^2+a^2}}]$
$\qquad=n[x+\sqrt{x^2+a^2}]^{n-1}\times \large\frac{\sqrt{x^2+a^2}+x)}{x^2+a^2}$
Step 3:
On simplifying we get,
$\Rightarrow \large\frac{n(x+\sqrt{(x^2+a^2)^n}}{\sqrt{x^2+a^2}}$
$y=[x+\sqrt{x^2+a^2}]^n$
$\Rightarrow \large\frac{ny}{\sqrt{x^2+a^2}}$
Hence proved.
answered Sep 30, 2013 by sreemathi.v
 

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