Use the chain rule to find #d/dx[tan(sqrtx)]# ?

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Answer 1 · 2018-01-11T03:38:49

#dy/dx= (sec^2sqrtx)/(2sqrtx)#

#y = tansqrtx#

Apply the chain rule and standard differential.

#dy/dx= sec^2sqrtx *d/dx sqrtx#

#dy/dx = sec^2sqrtx * 1/2x^(-1/2)#

#= (sec^2sqrtx)/(2sqrtx)#

Answer 2 · 2018-01-11T03:42:51

#d/dx[tan(sqrtx)]=sec^2(sqrtx)/(2sqrtx)#

Apply the chain rule:

#d/dx[f(g(x))]=f'(g(x))*g'(x)#

Let #f(x)=tan(x)# and #g(x)=sqrtx#

Thus, #f'(x)=sec^2x# and #g'(x)=1/(2sqrtx)#

So

#d/dx[tan(sqrtx)]=sec^2(sqrtx)*1/(2sqrtx)=sec^2(sqrtx)/(2sqrtx)#