Question

How do you find the derivative of `f(x)=pitan(pir^2-5r)`?

Answer 1

`(2\pi^2x-5\pi)sec^2(\pi x^2-5x)`

Explanation:

Assuming you meant `f(x)=\pi tan(\pi x^2-5x)`

`\pi` is a constant so we can ignore that. Since the inside of tangent is a function we have to use the chain rule. The derivative of `tan(x)` is `sec^2(x)` so we do get the following.
`\frac{d}{dx}f(x)=\pi sec^2(\pi x^2-5x)(\frac{d}{dx}g(x))`

Then we multiply the derivative of everything inside the parentheses which is
`\frac{d}{dx}g(x)=2\pir-5`

Combining them we get
`\frac{d}{dx}f(x)=\pi sec^2(\pi x^2-5x)(2\pix-5)=(2\pi^2x-5\pi)sec^2(\pi x^2-5x)`

You can expand it, but it might not look as clean.