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

May 22, 2018

$f ' \left(x\right) = \frac{9 x - x {e}^{x} - 2 {e}^{x}}{2 \sqrt{3 x - {e}^{x}}}$.

Explanation:

We have, $f \left(x\right) = x \sqrt{3 x - {e}^{x}}$.

Using the Product Rule, we get,

$\therefore f ' \left(x\right) = x \cdot \frac{d}{\mathrm{dx}} \left\{{\left(3 x - {e}^{x}\right)}^{\frac{1}{2}}\right\} + \sqrt{3 x - {e}^{x}} \cdot \frac{d}{\mathrm{dx}} \left\{x\right\} \ldots \ldots \left(\star\right)$.

Here, by the Chain Rule,

$\frac{d}{\mathrm{dx}} \left\{{\left(3 x - {e}^{x}\right)}^{\frac{1}{2}}\right\} = \frac{1}{2} \cdot {\left(3 x - {e}^{x}\right)}^{\frac{1}{2} - 1} \cdot \frac{d}{\mathrm{dx}} \left\{\left(3 x - {e}^{x}\right)\right\}$,

$= \frac{1}{2} \cdot {\left(3 x - {e}^{x}\right)}^{- \frac{1}{2}} \cdot \left[\frac{d}{\mathrm{dx}} \left\{3 x\right\} - \frac{d}{\mathrm{dx}} \left\{{e}^{x}\right\}\right]$,

$= \frac{1}{2 \sqrt{3 x - {e}^{x}}} \left\{3 \cdot 1 - {e}^{x}\right\}$.

$\Rightarrow \frac{d}{\mathrm{dx}} \left\{{\left(3 x - {e}^{x}\right)}^{\frac{1}{2}}\right\} = \frac{3 - {e}^{x}}{2 \sqrt{3 x - {e}^{x}}} \ldots \ldots \left({\star}_{1}\right)$.

Utilising $\left({\star}_{1}\right) \text{ in } \left(\star\right)$, we have,

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

$= \frac{\left(3 x - x {e}^{x}\right) + 2 \left(3 x - {e}^{x}\right)}{2 \sqrt{3 x - {e}^{x}}}$.

$\Rightarrow f ' \left(x\right) = \frac{9 x - x {e}^{x} - 2 {e}^{x}}{2 \sqrt{3 x - {e}^{x}}}$.

Enjoy Maths.!