How do you differentiate f(x)=sqrtsin((3x+7)^3) using the chain rule.?

Dec 21, 2015

Using chain rule consists in dealing with the separate operations in the equation one at a time.

Explanation:

First, chain rule states that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$, or, in this case, we'll use a longer one, following the same logic: $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dv}} \frac{\mathrm{dv}}{\mathrm{dw}} \frac{\mathrm{dw}}{\mathrm{dx}}$

Here, we can rename $y = \sqrt{u}$, $u = \sin \left(v\right)$, $v = {\left(w\right)}^{3}$ and $w = 3 x + 7$

Following the statement of the chain rule:

$\frac{\mathrm{dw}}{\mathrm{dx}} = 3$

$\frac{\mathrm{dv}}{\mathrm{dw}} = 3 {w}^{2}$

$\frac{\mathrm{du}}{\mathrm{dv}} = \cos \left(v\right)$

$\frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{2 \sqrt{u}}$

Aggregating them...

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{u}} \left(\cos \left(v\right)\right) \left(3 {w}^{2}\right) \left(3\right)$

Substituting $u$, $v$ and $w$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\cos \left({w}^{3}\right) \left(3 \left(3 x + 7\right)\right)}{2 \sqrt{\sin \left(v\right)}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\cos {\left(3 x + 7\right)}^{2} \left(9 x + 21\right)}{2 \sqrt{\sin \left({w}^{3}\right)}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\cos {\left(3 x + 7\right)}^{2} \left(9 x + 21\right)}{2 \sqrt{\sin {\left(3 x + 7\right)}^{3}}}$