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Differentiate the function with respect to x: $cos\;(sin x)$

$\begin{array}{1 1} -cosx.sin(sinx) \\ cosx.sin(sinx) \\ -sinx.sin(cosx) \\ -cosx.sin(cosx) \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(sinx)}{dx} $$= cosx$
  • $\; \large \frac{d(cosx)}{dx} $$= -sinx$
Given $y = cos (sinx)$
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(sinx)}{dx} $$= cosx$ and $\; \large \frac{d(cosx)}{dx} $$= -sinx$
$\Rightarrow$ Given $g(x) = sinx \rightarrow g'(x) = cosx$
$\Rightarrow f'(g(x)) = f'(cos (sinx)) = -sin(sinx)$
$\Rightarrow $ $y' = f'(g(x)).g'(x) = -sin(sinx).cosx = -cosx.sin(sinx)$
answered Apr 4, 2013 by balaji.thirumalai
 
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