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Differentiate the following w.r.t. \(x : \large \frac{e^x}{sin\;x} \\ \)

$\begin{array}{1 1} \large \frac{e^x(\sin x-\cos x)}{\sin^2x} \\ \large \frac{e^x(\sin x+\cos x)}{\sin^2x} \\ \large \frac{e^x(\cos x-\sin x)}{\sin^2x} \\ \large \frac{e^x(\cos x+\sin x)}{\sin^2x} \end{array} $

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Toolbox:
  • According to the Quotient Rule for differentiation, given two functions $u$ and $v,\; \large (\frac{u}{v})\;$$' = \large \frac{1}{v^2}$ $(v\;u' - u\;v')$
  • $\; \large \frac{de^x}{dx} $$=e^x $
Given $y = \large \frac{e^x}{sin\;x}$
This is of the form $\frac{u}{v}$ where $u = e^x$ and $v = \sin\;x$
According to the Quotient Rule for differentiation, given two functions $u$ and $v, \; \large \frac {d (\frac{u}{v})}{dx}$ $=\frac{1}{v^2}\; ( v\frac{du}{dx}-u \frac{dv}{dx})$
If $u = e^x, u' = du = e^x\;dx$
If $v = \sin x, v' = dv = \cos x\;dx$
$\Rightarrow$, Applying the quotient rule, $\large (\frac{u}{v})' = \frac{1}{(\sin x)^2}$$(\sin x\;e^x - e^x\;\cos x)$
$\Rightarrow \large (\frac{u}{v})'$$ = \large \frac{e^x(\sin x-\cos x)}{\sin^2x}$

 

answered Apr 10, 2013 by balaji.thirumalai
edited Apr 10, 2013 by balaji.thirumalai
 
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