Differential Calculus> differentiate the following function with...

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

taniska , 11 Years ago

Grade 12

2 Answers

Rinkoo Gupta

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

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Last Activity: 11 Years ago

Arun Kumar

Hello

$\\ \frac{d}{dx}\left(\cos ^x\left(x\right)+\sin ^{\frac{1}{x}}\left(x\right)\right) \\ \frac{d}{dx}\left(\cos ^x\left(x\right)\right) \\ =\frac{d}{du}\left(e^u\right)\frac{d}{dx}\left(x\ln \left(\cos \left(x\right)\right)\right) \\ \frac{d}{dx}\left(x\ln \left(\cos \left(x\right)\right)\right) \\ =\frac{d}{dx}\left(x\right)\ln \left(\cos \left(x\right)\right)+\frac{d}{dx}\left(\ln \left(\cos \left(x\right)\right)\right)x \\ =1\ln \left(\cos \left(x\right)\right)+\left(-\frac{\sin \left(x\right)}{\cos \left(x\right)}\right)x$

$\\ =e^u\left(\ln \left(\cos \left(x\right)\right)-\frac{x\sin \left(x\right)}{\cos \left(x\right)}\right) \\ =e^{x\ln \left(\cos \left(x\right)\right)}\left(\ln \left(\cos \left(x\right)\right)-\frac{x\sin \left(x\right)}{\cos \left(x\right)}\right) \\ =\cos ^x\left(x\right)\left(\ln \left(\cos \left(x\right)\right)-\frac{x\sin \left(x\right)}{\cos \left(x\right)}\right) \\ =\cos ^x\left(x\right)\left(\ln \left(\cos \left(x\right)\right)-\frac{x\sin \left(x\right)}{\cos \left(x\right)}\right)+\frac{\sin ^{\frac{1}{x}-1}\left(x\right)\left(x\cos \left(x\right)-\sin \left(x\right)\ln \left(\sin \left(x\right)\right)\right)}{x^2}$