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Differentiate the functions with respect to $x$: $sec \;( tan (\sqrt x ) )$

$\begin{array}{1 1} \frac{1}{2\sqrt x}.sec (tan\sqrt x). tan (sec\sqrt x) (sec^2\sqrt x) \\ \frac{-1}{2\sqrt x}.sec (tan\sqrt x). tan (sec\sqrt x) (sec^2\sqrt x) \\ \frac{1}{2\sqrt x}.sec (sec\sqrt x). tan (tan\sqrt x) (sec^2\sqrt x) \\ \frac{1}{2\sqrt x}.sec (sec\sqrt x). tan (sec\sqrt x) (sec^2\sqrt x) \end{array} $

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Toolbox:
  • According to the Chain Rule for differentiation, given two functions $f(x)$ and $g(x)$, and $y=f(g(x)) \rightarrow y' = f'(g(x)).g'(x)$.
  • $\; \large \frac{d(secx)}{dx} $$= secx.tanx$
  • $\; \large \frac{d(tanx)}{dx} $$= sec^2x$
Given $y = sec \;( tan (\sqrt x ) )$
According to the Chain Rule for differentiation, given two functions $f(x)$ and $g(x)$, and $y=f(g(x)) \rightarrow y' = f'(g(x)).g'(x)$
$\; \large \frac{d(secx)}{dx} $$= secx.tanx$ and $\; \large \frac{d(tanx)}{dx} $$= sec^2x$
$\textbf{Step 1}$:
Given $g(x) = tan (\sqrt x)$, we need to evaluate $g'(x)$
We apply the chain rule first on $g(x$), Let $g(x) = g(h(x))$, where $h(x) = \sqrt x$
$\Rightarrow$ Given $h(x) = \sqrt x \rightarrow h'(x) = \large \frac{1}{2\sqrt x}$
$\Rightarrow g'(h(x)) = g'(\sqrt x) = sec^2 \sqrt x$
$\Rightarrow $ $g'(x) = \large \frac{1}{2\sqrt x}.$$sec^2 \sqrt x$
$\textbf{Step 2}$:
$\Rightarrow f'(g(x)) = f'(tan \sqrt x) = sec (tan\sqrt x). tan (sec\sqrt x)$
$\Rightarrow $ $y' = f'(g(x)).g'(x) = sec (tan\sqrt x). tan (sec\sqrt x) \times \large \frac{1}{2\sqrt x}.$$sec^2 \sqrt x$
$\Rightarrow y' = \large \frac{1}{\sqrt2x}.$$sec (tan\sqrt x). tan (sec\sqrt x) (sec^2\sqrt x$)
answered Apr 4, 2013 by balaji.thirumalai
edited Jul 1, 2014 by rvidyagovindarajan_1
 
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