# differentiate the following function with respect to x(cos x)^x + (sin x)^1/x

Rinkoo Gupta
9 years ago

Y1=(cosx)x

taking log on both sides

logy1=xlogcosx

diff. using product rule with respect to x

(1/y1)dy1/dx=x.(1/cosx)(-sinx) +logcosx .1

dy1/dx=(cosx)x[-xtanx+logcosx]

y2=(sinx)1/x

taking log on both sides

logy2=1/x logsinx

diff. using product rule with respect to x

(1/y2)dy2/dx=(1/x)(1/sinx)(cosx)+logsinx(-1/x2)

dy2/dx=(sinx)1/x[(cotx)/x-(1/x2)logsinx]

hence dy/dx=(cosx)x[-xtanx+logcosx]+(sinx)1/x[(cotx)/x-(1/x2)logsinx]

Thanks & Regards

Rinkoo Gupta

Arun Kumar IIT Delhi
9 years ago
Hello
$\\ \frac{d}{dx}\left(\cos ^x\left(x\right)+\sin ^{\frac{1}{x}}\left(x\right)\right) \\ \frac{d}{dx}\left(\cos ^x\left(x\right)\right) \\ =\frac{d}{du}\left(e^u\right)\frac{d}{dx}\left(x\ln \left(\cos \left(x\right)\right)\right) \\ \frac{d}{dx}\left(x\ln \left(\cos \left(x\right)\right)\right) \\ =\frac{d}{dx}\left(x\right)\ln \left(\cos \left(x\right)\right)+\frac{d}{dx}\left(\ln \left(\cos \left(x\right)\right)\right)x \\ =1\ln \left(\cos \left(x\right)\right)+\left(-\frac{\sin \left(x\right)}{\cos \left(x\right)}\right)x$
$\\ =e^u\left(\ln \left(\cos \left(x\right)\right)-\frac{x\sin \left(x\right)}{\cos \left(x\right)}\right) \\ =e^{x\ln \left(\cos \left(x\right)\right)}\left(\ln \left(\cos \left(x\right)\right)-\frac{x\sin \left(x\right)}{\cos \left(x\right)}\right) \\ =\cos ^x\left(x\right)\left(\ln \left(\cos \left(x\right)\right)-\frac{x\sin \left(x\right)}{\cos \left(x\right)}\right) \\ =\cos ^x\left(x\right)\left(\ln \left(\cos \left(x\right)\right)-\frac{x\sin \left(x\right)}{\cos \left(x\right)}\right)+\frac{\sin ^{\frac{1}{x}-1}\left(x\right)\left(x\cos \left(x\right)-\sin \left(x\right)\ln \left(\sin \left(x\right)\right)\right)}{x^2}$
Thanks & Regards
Arun Kumar
Btech, IIT Delhi