Question #3063d

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Answer 1 · 2016-04-23T21:35:54

$5$

The chain rule states that

$d/dxf(g(x))=f'(g(x))*g'(x)$

Thus,

$d/dxf(e^tanx)=f'(e^tanx)*d/dx(e^tanx)$

$=f'(e^tanx) * e^tanx * sec^2x$

Note that the chain rule was also used to find $d/dx(e^tanx)$ .

So, if we want to find the derivative when $x=0$ , plug in $0$ for $x$ :

$d/dxf(e^tanx)|_(x=0)=f'(e^tan0) * e^tan0 * sec^2 0$

$=f'(e^0) * e^0 * (1)^2$

$=f'(1) * 1 * 1$

$=5$

Recall that $f'(1)=5$ was given in the question.