# differentiate the following function with respect to xe^sin x + (tan x)^x

Rinkoo Gupta
8 years ago

let y1=esinx

apply chain rule

dy/dx=esinx.cosx

let y2=tanxx

taking log on both sides

logy2=xlogtanx

diff.w.r.to x , we get

(1/y2)dy2/dx=x.(1/tanx)sec2x +logtanx.1

dy2/dx=tanxx[x/(sinxcosx)+logtanx]

hence dy/dx=esinx.cosx+tanxx[x/(sinxcosx)+logtanx]

Thanks & Regards

Rinkoo Gupta

$\\ \frac{d}{dx}\left(e^{\sin \left(x\right)}+\tan ^x\left(x\right)\right) \\ =\frac{d}{dx}\left(e^{\sin \left(x\right)}\right)+\frac{d}{dx}\left(\tan ^x\left(x\right)\right) \\ \frac{d}{dx}\left(e^{\sin \left(x\right)}\right) \\ =\frac{d}{du}\left(e^u\right)\frac{d}{dx}\left(\sin \left(x\right)\right) \\ \frac{d}{dx}\left(\sin \left(x\right)\right) \\ =\cos \left(x\right) \\ =e^{\sin \left(x\right)}\cos \left(x\right) \\ \frac{d}{dx}\left(\tan ^x\left(x\right)\right)$
$\\ =\frac{d}{du}\left(e^u\right)\frac{d}{dx}\left(x\ln \left(\tan \left(x\right)\right)\right) \\ \frac{d}{dx}\left(x\ln \left(\tan \left(x\right)\right)\right) \\ =\frac{d}{dx}\left(x\right)\ln \left(\tan \left(x\right)\right)+\frac{d}{dx}\left(\ln \left(\tan \left(x\right)\right)\right)x \\ =\frac{x}{\cos ^2\left(x\right)\tan \left(x\right)}+\ln \left(\tan \left(x\right)\right) \\ =e^{x\ln \left(\tan \left(x\right)\right)}\left(\frac{x}{\cos ^2\left(x\right)\tan \left(x\right)}+\ln \left(\tan \left(x\right)\right)\right) \\ =e^{\sin \left(x\right)}\cos \left(x\right)+\tan ^x\left(x\right)\left(\frac{x}{\cos ^2\left(x\right)\tan \left(x\right)}+\ln \left(\tan \left(x\right)\right)\right)$