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Question Number 83156 by 09658867628 last updated on 28/Feb/20
find the derivtive of y=e^(cos x)
$${find}\:{the}\:{derivtive}\:{of}\:{y}={e}^{\mathrm{cos}\:{x}} \\ $$
Commented by niroj last updated on 28/Feb/20
   y= e^(cos x)     D.w.r.to x.    (dy/dx)= e^(cos x) (−sin x)     (dy/dx)= −sin x .e^(cos x)
$$\:\:\:\mathrm{y}=\:\mathrm{e}^{\mathrm{cos}\:\mathrm{x}} \\ $$$$\:\:\mathrm{D}.\mathrm{w}.\mathrm{r}.\mathrm{to}\:\mathrm{x}. \\ $$$$\:\:\frac{\mathrm{dy}}{\mathrm{dx}}=\:\mathrm{e}^{\mathrm{cos}\:\mathrm{x}} \left(−\mathrm{sin}\:\mathrm{x}\right) \\ $$$$\:\:\:\frac{\mathrm{dy}}{\mathrm{dx}}=\:−\mathrm{sin}\:\mathrm{x}\:.\mathrm{e}^{\mathrm{cos}\:\mathrm{x}} \\ $$
Commented by peter frank last updated on 28/Feb/20
lny=lne^(cos x)   lny=cos xlne  lny=cos x  (y^′ /y)=sin  x  y^′ =ysin  x  y^′ =−e^(cos x) sin x
$${lny}={lne}^{\mathrm{cos}\:{x}} \\ $$$${lny}=\mathrm{cos}\:{xlne} \\ $$$${lny}=\mathrm{cos}\:{x} \\ $$$$\frac{{y}^{'} }{{y}}=\mathrm{sin}\:\:{x} \\ $$$${y}^{'} ={y}\mathrm{sin}\:\:{x} \\ $$$${y}^{'} =−{e}^{\mathrm{cos}\:{x}} \mathrm{sin}\:{x} \\ $$
Answered by jagoll last updated on 28/Feb/20
y ′ = −sin (x) e^(cos (x))
$$\mathrm{y}\:'\:=\:−\mathrm{sin}\:\left(\mathrm{x}\right)\:\mathrm{e}^{\mathrm{cos}\:\left(\mathrm{x}\right)} \: \\ $$

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