Question

How do you differentiate `r=5theta^2sectheta`?

Answer 1

`(dr)/(d theta)=5thetasectheta(tantheta+2)`

Explanation:

differentiate using the `color(blue)"product rule"`

`"Given " r=f(theta).g(theta)" then"`

`color(red)(bar(ul(|color(white)(2/2)color(black)(r'(theta)=f(theta)g'(theta)+g(theta)f'(theta))color(white)(2/2)|)))larr" product rule"`

The standard derivative of `sec(theta)` which should be known is.

`color(red)(bar(ul(|color(white)(2/2)color(black)(d/(d theta)(sec(theta))=sec(theta)tan(theta))color(white)(2/2)|)))`

`"here " f(theta)=5theta^2rArrf'(theta)=10theta`

`"and " g(theta)=secthetarArrg'(theta)=secthetatantheta`

`rArrr'(theta)=5theta^2secthetatantheta+sectheta.10theta`

`=5thetasectheta(tantheta+2)`