# Question #aaeae

May 26, 2017

a) $3 {x}^{2} - \frac{8}{{x}^{3}} + \frac{1}{2 \sqrt{x}}$
b) $40 {\left(2 x - 1\right)}^{3}$
c)$\frac{2 x}{\sin} x - \frac{{x}^{2} \cos x}{\sin} ^ 2 x - \cos \frac{x}{\sin} ^ 2 x$

#### Explanation:

a) $\frac{d}{\mathrm{dx}} \left[{x}^{3} + \frac{4}{x} ^ 2 + {x}^{\frac{1}{2}}\right] = \frac{d}{\mathrm{dx}} \left[{x}^{3}\right] + 4 \frac{d}{\mathrm{dx}} \left[{x}^{- 2}\right] + \frac{d}{\mathrm{dx}} \left[{x}^{\frac{1}{2}}\right]$

Using the power rule:

$= 3 {x}^{2} - 8 {x}^{-} 3 + \frac{1}{2} {x}^{- \frac{1}{2}}$
$= 3 {x}^{2} - \frac{8}{{x}^{3}} + \frac{1}{2 \sqrt{x}}$

b) $\frac{d}{\mathrm{dx}} \left[5 {\left(2 x - 1\right)}^{4}\right] = 5 \frac{d}{\mathrm{dx}} \left[{\left(2 x - 1\right)}^{2}\right]$

The chain rule states that:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

Let $y = {u}^{4}$ while $u = 2 x - 1$

Using the power rule:

$\frac{\mathrm{dy}}{\mathrm{du}} = 4 {g}^{3} , \setminus \quad \frac{\mathrm{du}}{\mathrm{dx}} = 2$

$5 \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}} = 40 {\left(2 x - 1\right)}^{3}$

c) $\frac{d}{\mathrm{dx}} \left[\frac{{x}^{2} + 4}{\sin} x\right] = \frac{d}{\mathrm{dx}} \left[{x}^{2} / \sin x\right] + 4 \frac{d}{\mathrm{dx}} \left[\frac{1}{\sin} x\right]$

Solve for $\frac{d}{\mathrm{dx}} \left[{x}^{2} / \sin x\right]$ first.

Using the product rule:

$\frac{d}{\mathrm{dx}} \left[f \left(x\right) \cdot g \left(x\right)\right] = f \left(x\right) \frac{\mathrm{dg}}{\mathrm{dx}} + g \left(x\right) \frac{\mathrm{df}}{\mathrm{dx}}$

$\frac{d}{\mathrm{dx}} \left[{x}^{2} \cdot \frac{1}{\sin} x\right] = \frac{d}{\mathrm{dx}} \left[{x}^{2}\right] \cdot \frac{1}{\sin} x + \frac{d}{\mathrm{dx}} \left[\frac{1}{\sin} x\right] \cdot {x}^{2}$
$= \frac{2 x}{\sin} x + \frac{d}{\mathrm{dx}} \left[\frac{1}{\sin} x\right] \cdot {x}^{2}$

Solve for $\frac{d}{\mathrm{dx}} \left[\frac{1}{\sin} x\right]$

Using the chain rule, let:
$y = \frac{1}{u}$ while $u = \sin x$
$\frac{\mathrm{dy}}{\mathrm{du}} = - \frac{1}{u} ^ 2$

$\frac{\mathrm{du}}{\mathrm{dx}} = \cos x$

$\frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}} = - \cos \frac{x}{u} ^ 2$

Sub back $u = \sin x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \cos \frac{x}{\sin} ^ 2 x$

We can also use the reciprocal rule to solve for $\frac{1}{\sin} x$

$\frac{d}{\mathrm{dx}} \left[\frac{1}{f} \left(x\right)\right] = - \frac{\mathrm{df}}{\mathrm{dx}} \cdot \frac{1}{f} ^ 2 \left(x\right)$

Finally, $\frac{d}{\mathrm{dx}} \left[{x}^{2} / \sin x\right] = \frac{2 x}{\sin} x - \frac{{x}^{2} \cos x}{\sin} ^ 2 x$

Now solve for part 2
$\frac{d}{\mathrm{dx}} \left[\frac{1}{\sin} x\right]$ has already been solved

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

Now, sub back in.

$\frac{d}{\mathrm{dx}} \left[{x}^{2} / \sin x\right] + 4 \frac{d}{\mathrm{dx}} \left[\frac{1}{\sin} x\right] = \frac{2 x}{\sin} x - \frac{{x}^{2} \cos x}{\sin} ^ 2 x - \cos \frac{x}{\sin} ^ 2 x$