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

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Answer 1 · 2016-01-09T16:08:36

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

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 $1/(2sqrtx)$ .

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

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