Browse Questions

Differentiate the following w.r.t. $x:$ $cos \: ( log\: x + e^x ), x > 0$

$\begin{array}{1 1} -sin (\log x + e^x)\; (\large \frac{1}{x}+ e^x) \\sin (\log x + e^x)\; (\large \frac{1}{x}- e^x) \\ sin (\log x + e^x)\; (\large \frac{1}{x}+ e^x) \\-sin (\log x + e^x)\; (\large \frac{1}{x}- e^x) \end{array}$

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(cosx)}{dx} $$=-sinx • \; \large \frac{d(e^{x})}{dx}$$= e^{x}$
• $\; \large \frac{d(\log x)}{dx} $$=\large\frac{1}{x} Given y = cos \: ( log\: x + e^x ), x > 0 This is of the form y = f(g(x), where g(x) =\log x + e^x ,\; x > 0, so we can apply the chain rule of differentiation. We know that \; \large \frac{d(e^{x})}{dx}$$= e^{x}$ and $\; \large \frac{d(\log x)}{dx} $$=\large\frac{1}{x} \Rightarrow g'(x) = \large \frac{1}{x}$$+ e^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(cosx)}{dx} $$=-sinx \Rightarrow y' = -sin (\log x + e^x)\; (\large \frac{1}{x}$$+ e^x)$
edited Apr 10, 2013