How do you differentiate #sqrtx(sinx+cosx)#?

1 Answer
Harish Chandra Rajpoot
Jul 19, 2018

#(\frac{1-2x}{2\sqrtx})\sin x+(\frac{1+2x}{2\sqrtx})\cos x#

Explanation:

Differentiating given function: #f(x)=\sqrtx(\sinx +\cosx )# w.r.t. #x# using product rule as follows

#f'(x)=d/dx(\sqrtx(\sinx +\cosx ))#

#=\sqrtxd/dx(\sin x+\cos x)+(\sin x+\cos x)d/dx\sqrtx#

#=\sqrtx(\cos x-\sin x)+(\sin x+\cos x)1/{2\sqrtx}#

#=(\frac{1-2x}{2\sqrtx})\sin x+(\frac{1+2x}{2\sqrtx})\cos x#

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