# How do you differentiate f(x) = 3x(e^((x-9)^2))^3 using the chain rule?

Jun 1, 2017

${e}^{{\left(x - 9\right)}^{6}} \left(18 x {\left(x - 9\right)}^{5} + 3\right)$

#### Explanation:

Step 1. Simplify the exponent

$f \left(x\right) = 3 x {e}^{{\left(x - 9\right)}^{6}}$

Step 2. Using the Product Rule directly uses the chain rule.

$f ' \left(x\right) = 3 x \frac{d}{\mathrm{dx}} \left({e}^{{\left(x - 9\right)}^{6}}\right) + {e}^{{\left(x - 9\right)}^{6}} \frac{d}{\mathrm{dx}} \left(3 x\right)$

$= 3 x {e}^{{\left(x - 9\right)}^{6}} \frac{d}{\mathrm{dx}} \left({\left(x - 9\right)}^{6}\right) + 3 {e}^{{\left(x - 9\right)}^{6}}$

$= 3 x {e}^{{\left(x - 9\right)}^{6}} \left(6 {\left(x - 9\right)}^{5}\right) + 3 {e}^{{\left(x - 9\right)}^{6}}$

$= 18 x {\left(x - 9\right)}^{5} {e}^{{\left(x - 9\right)}^{6}} + 3 {e}^{{\left(x - 9\right)}^{6}}$

Step 3. Factor out the common ${e}^{{\left(x - 9\right)}^{6}}$ term

$= {e}^{{\left(x - 9\right)}^{6}} \left(18 x {\left(x - 9\right)}^{5} + 3\right)$