How do you differentiate  f(x)=(1-e^(3sqrtx))^2 using the chain rule.?

Jan 25, 2018

$f ' \left(x\right) = \frac{3 {e}^{6 \sqrt{x}}}{\sqrt{x}} \left(1 - {e}^{- 3 \sqrt{x}}\right) .$

Explanation:

Note that $f \left(x\right)$ is a compound function envolving the functions $g \left(x\right) = {x}^{2}$ and $h \left(x\right) = 1 - {e}^{x}$ and $p \left(x\right) = 3 \sqrt{x}$. In this sense, we could write that $f \left(x\right) = g \left(h \left(p \left(x\right)\right)\right) .$

We want to find the derivative $\frac{\mathrm{df}}{\mathrm{dx}}$ of the function $f \left(x\right) = g \left(h \left(x\right)\right) .$ Using the chain rule:

$f ' \left(x\right) = g ' \left[\left(h \left(p \left(x\right)\right)\right)\right] \cdot h ' \left[\left(p \left(x\right)\right)\right] \cdot p ' \left(x\right)$.

Let us calculate each term separately.

1) $g ' \left[\left(h \left(p \left(x\right)\right)\right)\right] = 2 \left(1 - {e}^{3 \sqrt{x}}\right)$;

2) $h ' \left[\left(p \left(x\right)\right)\right] = - {e}^{3 \sqrt{x}}$;

3) $p ' \left(x\right) = \frac{3}{2} {x}^{- \frac{1}{2}}$.

Then:

$f ' \left(x\right) = \cancel{2} \left(1 - {e}^{3 \sqrt{x}}\right) \cdot \left(- {e}^{3 \sqrt{x}}\right) \cdot \left(\frac{3}{\cancel{2}} {x}^{- \frac{1}{2}}\right)$;

$f ' \left(x\right) = \frac{3}{\sqrt{x}} {e}^{3 \sqrt{x}} \left({e}^{3 \sqrt{x}} - 1\right)$;

$f ' \left(x\right) = \frac{3 {e}^{6 \sqrt{x}}}{\sqrt{x}} \left(1 - {e}^{- 3 \sqrt{x}}\right) .$

Hope it helped you!