# Question 5f32d

Feb 15, 2018

$- \cot x \csc x$

#### Explanation:

We have $\frac{d}{\mathrm{dx}} \frac{1}{\sin} x$

We can apply the quotient rule, that is, $\left(\frac{f}{g}\right) '$, where Newton's notation is used and $f$ and $g$ are functions, is equal to $\frac{f ' g - f g '}{g} ^ 2$.

Here, $f = 1$ and $g = \sin x$. We can input:

$\frac{\frac{d}{\mathrm{dx}} \left(1\right) \cdot \sin x - 1 \cdot \frac{d}{\mathrm{dx}} \left(\sin x\right)}{\sin} ^ 2 x$

We must find $f '$ and $g '$.

$\frac{d}{\mathrm{dx}} 1 = 0$, so the above equation reduces to:

$\frac{- \frac{d}{\mathrm{dx}} \sin x}{\sin} ^ 2 x$

Since $\frac{d}{\mathrm{dx}} \sin x = \cos x$, we can input:

$- \cos \frac{x}{\sin} ^ 2 x$

Or:

$- \cos \frac{x}{\sin} x \cdot \frac{1}{\sin} x$

Since $\cos \frac{x}{\sin} x = \cot x$ and $\frac{1}{\sin} x = \csc x$, this becomes:

$- \cot x \csc x$

Feb 15, 2018

$- \cot x \csc x$

#### Explanation:

$\text{differentiate using the "color(blue)"chain rule}$

$\text{given "y=f(g(x))" then}$

dy/dx=f'(g(x)xxg'(x)

$\text{here } y = \frac{1}{\sin} x = {\left(\sin x\right)}^{-} 1$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = - {\left(\sin x\right)}^{-} 2 \times \frac{d}{\mathrm{dx}} \left(\sin x\right)$

$\textcolor{w h i t e}{\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}} = - \cos x {\left(\sin x\right)}^{-} 2$

color(white)(rArrdy/dx)=-cosx/(sin^2x#

$\textcolor{w h i t e}{\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}} = - \cos \frac{x}{\sin} x \times \frac{1}{\sin} x$

$\textcolor{w h i t e}{\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}} = - \cot x \csc x$