How do you differentiate $f(x)=sec(e^(sqrtx-4) )$ using the chain rule?

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Answer 1 · 2016-01-23T04:02:41

$f'(x)=(e^(sqrtx-4)sec(e^(sqrtx-4))tan(e^(sqrtx-4)))/(2sqrtx)$

The chain rule states that $d/dx[sec(u)]=u'*sec(u)tan(u)$ , and here $u=e^(sqrtx-4)$ .

$f'(x)=d/dx[e^(sqrtx-4)]*sec(e^(sqrtx-4))tan(e^(sqrtx-4))$

To find this derivative, use the rule that $d/dx[e^u]=u'*e^u$ , and this time $u=sqrtx-4$ .

$f'(x)=d/dx[sqrtx-4]*e^(sqrtx-4)sec(e^(sqrtx-4))tan(e^(sqrtx-4))$

To find $d/dx[sqrtx-4]$ , use the power rule, in that $d/dx[x^(1/2)-4]=1/2x^(-1/2)=1/(2sqrtx)$ . This gives

$f'(x)=(e^(sqrtx-4)sec(e^(sqrtx-4))tan(e^(sqrtx-4)))/(2sqrtx)$

How do you differentiate f(x)=sec(e^(sqrtx-4) ) f(x)=sec(e^(sqrtx-4) ) using the chain rule?