How do you differentiate $f (x) = e^{sec \sqrt{x}}$ $f(x)=e^(secsqrtx)$ using the chain rule.?

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Answer 1 · 2015-12-11T00:26:44

$f'(x)=(sec(sqrtx)tan(sqrtx)e^(sec(sqrtx)))/(2sqrtx)$

According to the chain rule,

$f'(x)=e^(secsqrtx)d/dx[secsqrtx]$

First, find the internal derivative (again using the chain rule).

$d/dx[secsqrtx]=secsqrtxtansqrtxd/dx[sqrtx]$

$=1/2x^(-1/2)secsqrtxtansqrtx$

$=(secsqrtxtansqrtx)/(2sqrtx)$

Plug this back in to find $f'(x)$ .

$f'(x)=(e^(secsqrtx))(secsqrtxtansqrtx)/(2sqrtx)$

$f'(x)=(sec(sqrtx)tan(sqrtx)e^(sec(sqrtx)))/(2sqrtx)$

How do you differentiate f(x)=e^(secsqrtx)f(x)=esec√xf(x)=e^(secsqrtx) using the chain rule.?