How do you differentiate f(x)=(x/sin(7x))^2 using the chain rule?

Jan 17, 2016

$f ' \left(x\right) = \frac{2 x \left(\sin \left(7 x\right) - 7 x \cos \left(7 x\right)\right)}{\sin} ^ 3 \left(7 x\right)$

Explanation:

First issue: the power of $2$.

$f ' \left(x\right) = \frac{2 x}{\sin \left(7 x\right)} \cdot \frac{d}{\mathrm{dx}} \left[\frac{x}{\sin} \left(7 x\right)\right]$

Now, use quotient rule to find the internal derivative.

$\frac{d}{\mathrm{dx}} \left[\frac{x}{\sin} \left(7 x\right)\right] = \frac{\sin \left(7 x\right) \frac{d}{\mathrm{dx}} \left[x\right] - x \frac{d}{\mathrm{dx}} \left[\sin \left(7 x\right)\right]}{\sin} ^ 2 \left(7 x\right)$

Note that finding $\frac{d}{\mathrm{dx}} \left[\sin \left(7 x\right)\right]$ will require the chain rule, insofar as $\frac{d}{\mathrm{dx}} \left[\sin \left(u\right)\right] = \cos \left(u\right) \cdot \frac{\mathrm{du}}{\mathrm{dx}}$.

$\frac{d}{\mathrm{dx}} \left[\frac{x}{\sin} \left(7 x\right)\right] = \frac{\sin \left(7 x\right) \times 1 - x \left(\cos \left(7 x\right) \times 7\right)}{\sin} ^ 2 \left(7 x\right)$

Put this back into the original expression.

$f ' \left(x\right) = \frac{2 x}{\sin} \left(7 x\right) \cdot \frac{\sin \left(7 x\right) - 7 x \cos \left(7 x\right)}{\sin} ^ 2 \left(7 x\right)$

$f ' \left(x\right) = \frac{2 x \left(\sin \left(7 x\right) - 7 x \cos \left(7 x\right)\right)}{\sin} ^ 3 \left(7 x\right)$