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Differentiate with respect to x \( \sec(\tan(\sqrt x))\)

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  • $\; \large \frac{d(\sec x)}{dx} $$= \sec x.\tan x$
  • $\; \large \frac{d(\tan x)}{dx} $$= \sec^2x$
Given :
$y=\sec(\tan \sqrt x)$
Differentiating with respect to $x$ we get,
$\large\frac{1}{2\sqrt x}=\frac{du}{dx}$
Let $\tan u=v$
Diff w.r.t $u$ we get,
Step 2:
Let $y=\sec v$
Diff w.r.t $v$ we get,
$\large\frac{dy}{dv}$$=\sec v\tan v$
$\large\frac{dy}{dx}=\frac{dy}{dv}\times \frac{du}{dx}$
Step 3:
Now substituting the values we get,
$\large\frac{dy}{dx}$$=\sec v.\tan v\times \sec^2 u\times \large\frac{1}{2\sqrt x}$
Substituting back for $u$ and $v$ we get,
$(\sec(\tan \sqrt x)\tan (\tan \sqrt x)\times \sec^2(\sqrt x)\times \large\frac{1}{2\sqrt x}$
Hence $\large\frac{dy}{dx}=\frac{\sec(\tan \sqrt x)\tan(\tan \sqrt x).\sec^2 \sqrt x}{2\sqrt x}$
answered Sep 24, 2013 by sreemathi.v
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