How do you differentiate #sqrtx(sinx+cosx)#? Calculus Basic Differentiation Rules Power Rule 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# Answer link Related questions How do you find the derivative of a polynomial? How do you find the derivative of #y =1/sqrt(x)#? How do you find the derivative of #y =4/sqrt(x)#? How do you find the derivative of #y =sqrt(2x)#? How do you find the derivative of #y =sqrt(3x)#? How do you find the derivative of #y =sqrt(x)#? How do you find the derivative of #y =sqrt(x)# using the definition of derivative? How do you find the derivative of #y =sqrt(3x+1)#? How do you find the derivative of #y =sqrt(9-x)#? How do you find the derivative of #y =sqrt(x-1)#? See all questions in Power Rule Impact of this question 1659 views around the world You can reuse this answer Creative Commons License