How do you differentiate #f(x)=e^(cotsqrtx)# using the chain rule.?

1 Answer
mason m
Jan 9, 2016

#f'(x)=(-csc^2sqrtx*e^cotsqrtx)/(2sqrtx)#

Explanation:

First Issue: The #e# power. According to the chain rule, #d/dx(e^u)=e^u*u'#. Thus,

#f'(x)=e^cotsqrtx*d/dx(cotsqrtx)#

Second Issue: the cotangent function. Again through the chain rule, #d/dx(cotu)=-csc^2u*u'#. Thus,

#f'(x)=e^cotsqrtx*-csc^2sqrtx*d/dx(sqrtx)#

Now, to differentiate #sqrtx#, treat it as #x^(1/2)#. Thus, its derivative is #1/2x^(-1/2)# or #1/(2sqrtx)#.

#f'(x)=(-csc^2sqrtx*e^cotsqrtx)/(2sqrtx)#

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