How do you differentiate f(x)=e^sqrt(1-(3x+5)^2) using the chain rule.?

Jan 7, 2016

dy/dx = -(3e^(sqrt(1-(3x+5)^2))*(3x+5))/(sqrt(1-(3x+5)^2)

Explanation:

$y = {e}^{\sqrt{1 - {\left(3 x + 5\right)}^{2}}}$

Say that $\sqrt{1 - {\left(3 x + 5\right)}^{2}} = u$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} {e}^{u} \frac{\mathrm{du}}{\mathrm{dx}}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{u} \frac{d}{\mathrm{dx}} \sqrt{1 - {\left(3 x + 5\right)}^{2}}$

Say that $v = 1 - {\left(3 x + 5\right)}^{2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{u} \frac{d}{\mathrm{dx}} \sqrt{v} \frac{\mathrm{dv}}{\mathrm{dx}}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{u} / \left(2 \sqrt{v}\right) \frac{d}{\mathrm{dx}} \left(1 - {\left(3 x + 5\right)}^{2}\right)$

Say that $3 x + 5 = w$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{u} / \left(2 \sqrt{v}\right) \frac{d}{\mathrm{dx}} \left(1 - {w}^{2}\right) \frac{\mathrm{dw}}{\mathrm{dx}}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{{e}^{u} w}{\sqrt{v}} \frac{d}{\mathrm{dx}} \left(3 x + 5\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{3 {e}^{u} w}{\sqrt{v}}$

Now put it all back in terms of $x$

dy/dx = -(3e^(sqrt(1-(3x+5)^2))*(3x+5))/(sqrt(1-(3x+5)^2)