# Question #3063d

Apr 23, 2016

$5$

#### Explanation:

The chain rule states that

$\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

Thus,

$\frac{d}{\mathrm{dx}} f \left({e}^{\tan} x\right) = f ' \left({e}^{\tan} x\right) \cdot \frac{d}{\mathrm{dx}} \left({e}^{\tan} x\right)$

$= f ' \left({e}^{\tan} x\right) \cdot {e}^{\tan} x \cdot {\sec}^{2} x$

Note that the chain rule was also used to find $\frac{d}{\mathrm{dx}} \left({e}^{\tan} x\right)$.

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

$\frac{d}{\mathrm{dx}} f \left({e}^{\tan} x\right) {|}_{x = 0} = f ' \left({e}^{\tan} 0\right) \cdot {e}^{\tan} 0 \cdot {\sec}^{2} 0$

$= f ' \left({e}^{0}\right) \cdot {e}^{0} \cdot {\left(1\right)}^{2}$

$= f ' \left(1\right) \cdot 1 \cdot 1$

$= 5$

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