How do you differentiate #f(x)=xsinsqrtx# using the chain rule?

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Answer 1 · 2017-03-27T23:34:13

#sinsqrtx + (cossqrtx * sqrtx)/2#

In order to differentiate #f(x) = xsinsqrtx#, we need to use the product rule

Product rule: #f(x)*g(x) = f'(x)g(x) + g'(x)f(x)#

#f(x)*g(x) = f'(x)g(x) + g'(x)f(x)#
#x*sinsqrtx = d/dx x * sinsqrtx + d/dx sinsqrtx * x#

#x' = 1#
#sinsqrtx' = cossqrtx * d/dx sqrtx#

#1 * sinsqrtx + cossqrtx * d/dx sqrtx * x#
#sqrtx' = x^(1/2)' = (1/2)*x^((1/2) -1)= 1/(2sqrtx)#

#sinsqrtx + cossqrtx * 1/(2sqrtx) * x#
#sinsqrtx + (cossqrtx * sqrtx)/2#