Differential Calculus> differentiate the following function with...

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

taniska , 11 Years ago

Grade 12

2 Answers

Rinkoo Gupta

let y₁=e^sinx

^{apply chain rule}

dy/dx=e^sinx.cosx

let y₂=tanx^x

^{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=tanx^x[x/(sinxcosx)+logtanx]

hence dy/dx=e^sinx.cosx+tanx^x[x/(sinxcosx)+logtanx]

Thanks & Regards

Rinkoo Gupta

AskIITians Faculty

Last Activity: 11 Years ago

Arun Kumar

Hello

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

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