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How do you differentiate #f(x)=tan(sqrt(x^2-3)) # using the chain rule?

1 Answer

Jan 14, 2016

#f'(x)=(xsec^2(sqrt(x^2-3)))/sqrt(x^2-3)#

Explanation:

We may use the following rules in conjunction :

#d/dxtan[u(x)]=sec^2u*(du)/dx#

#d/dx[u(x)]^n=n[u(x)]^(n-1)*(du)/dx#

Hence we obtain :

#f'(x)=sec^2(sqrt(x^2-3)) * 1/2(x^2-3)^(-1/2) *(2x) #

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