Question

find $\frac{dp}{dq}$ for $p = \frac{cos q-sin q}{cos q}$. $\frac{dp}{dq}=square$

Explanation:

Step1: Simplify the function

First, rewrite $p=\frac{\cos q-\sin q}{\cos q}=1 - \tan q$.

Step2: Differentiate term - by - term

The derivative of a constant is 0, and the derivative of $\tan q$ is $\sec^{2}q$. Using the difference rule of differentiation $\frac{d}{dq}(u - v)=\frac{du}{dq}-\frac{dv}{dq}$, where $u = 1$ and $v=\tan q$. So $\frac{dp}{dq}=\frac{d}{dq}(1)-\frac{d}{dq}(\tan q)$.

Step3: Calculate the derivatives

Since $\frac{d}{dq}(1) = 0$ and $\frac{d}{dq}(\tan q)=\sec^{2}q$, we have $\frac{dp}{dq}=0-\sec^{2}q=-\sec^{2}q$.

Answer:

$-\sec^{2}q$