# How do you differentiate  f(x)=(4+e^(sqrt(7x)))^3 using the chain rule.?

$f ' \left(x\right) = \frac{21}{2 \setminus \sqrt{7 x}} {\left(4 + {e}^{\setminus \sqrt{7 x}}\right)}^{2} {e}^{\setminus \sqrt{7 x}}$

#### Explanation:

Given function:

$f \left(x\right) = {\left(4 + {e}^{\setminus \sqrt{7 x}}\right)}^{3}$

differentiating above equation w.r.t. $x$ using chain rule as folows

$\frac{d}{\mathrm{dx}} f \left(x\right) = \frac{d}{\mathrm{dx}} {\left(4 + {e}^{\setminus \sqrt{7 x}}\right)}^{3}$

$f ' \left(x\right) = 3 {\left(4 + {e}^{\setminus \sqrt{7 x}}\right)}^{2} \frac{d}{\mathrm{dx}} \left(4 + {e}^{\setminus \sqrt{7 x}}\right)$

$= 3 {\left(4 + {e}^{\setminus \sqrt{7 x}}\right)}^{2} \left(0 + {e}^{\setminus \sqrt{7 x}} \frac{d}{\mathrm{dx}} \left(\setminus \sqrt{7 x}\right)\right)$

$= 3 {\left(4 + {e}^{\setminus \sqrt{7 x}}\right)}^{2} \left({e}^{\setminus \sqrt{7 x}} \left(\frac{1}{2 \setminus \sqrt{7 x}} \frac{d}{\mathrm{dx}} \left(7 x\right)\right)\right)$

=3(4+e^{\sqrt{7x}})^2({e^{\sqrt{7x}}}/{2\sqrt{7x}}(7)))

$= \frac{21}{2 \setminus \sqrt{7 x}} {\left(4 + {e}^{\setminus \sqrt{7 x}}\right)}^{2} {e}^{\setminus \sqrt{7 x}}$