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Differentiate the functions with respect to x in $\;\cos(\sqrt {x} \;) $

$\begin{array}{1 1} \large \frac{-sin \sqrt x}{2 \sqrt x} \\ \large \frac{cos \sqrt x}{2 \sqrt x} \\ \large \frac{-sin \sqrt x}{ \sqrt x} \\ \large \frac{cos \sqrt x}{ \sqrt x}\end{array} $

1 Answer

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\sqrt x}{dx}$$ = \large \frac{1}{2 \sqrt x}$
  • $\; \large \frac{d(cosx)}{dx} $$=-sinx $
Given $y = \;\cos \sqrt {x} $. This is of the form $y = f(g(x))$ where $g(x) = \sqrt x$.
We can apply the Chain rule. 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\sqrt x}{dx}$$ = \large \frac{1}{2 \sqrt x}$
$\Rightarrow g'(x) = \large \frac{1}{2 \sqrt x}$
$\; \large \frac{d(cosx)}{dx} $$=-sinx $
$\Rightarrow f'(g(x)) = -sin \sqrt x$
$\Rightarrow y' = f'(g(x)).g'(x) = -sin \sqrt x. \large \frac{1}{2 \sqrt x} $$= \large \frac{-sin \sqrt x}{2 \sqrt x}$
answered Apr 5, 2013 by balaji.thirumalai
 
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