Question

find $\frac{dp}{dq}$. $p = 3+\frac{1}{sin q}$ $\frac{dp}{dq}=$

Explanation:

Step1: Rewrite the function

Rewrite $p = 3+\frac{1}{\sin q}$ as $p = 3+(\sin q)^{- 1}$.

Step2: Differentiate term - by - term

The derivative of a constant is 0, so $\frac{d}{dq}(3)=0$. For the second term, use the chain - rule. If $u = \sin q$, then $y = u^{-1}$. The derivative of $y$ with respect to $u$ is $\frac{dy}{du}=-u^{-2}=-\frac{1}{u^{2}}$, and the derivative of $u=\sin q$ with respect to $q$ is $\frac{du}{dq}=\cos q$. By the chain - rule $\frac{dy}{dq}=\frac{dy}{du}\cdot\frac{du}{dq}$.

Step3: Apply the chain - rule result

$\frac{d}{dq}(\sin q)^{-1}=-\frac{1}{(\sin q)^{2}}\cdot\cos q=-\frac{\cos q}{\sin^{2}q}$.

Step4: Combine the derivatives of the terms

Since $\frac{d}{dq}(3) = 0$ and $\frac{d}{dq}(\sin q)^{-1}=-\frac{\cos q}{\sin^{2}q}$, then $\frac{dp}{dq}=0-\frac{\cos q}{\sin^{2}q}=-\frac{\cos q}{\sin^{2}q}$.

Answer:

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