# Thread: Simplifying a denominator before taking derivative

1. ## Simplifying a denominator before taking derivative

This might be more of a trig question, so if it is would someone kindly move it for me?

So I'm finding the second derivative of $ln[sec(x)+tan(x)]$. My first derivative is $\frac{sec(x)tan(x)+sec^2(x)}{sec(x)+tan(x)}$.

My question is: can I simplify this fraction to make taking the second derivative less tedious? I'm capable of doing it without simplifying first, but I don't want to get a super long and ugly answer.

2. ## Re: Simplifying a denominator before taking derivative

Have you considered factoring out sec(x) from the numerator?

3. ## Re: Simplifying a denominator before taking derivative

Originally Posted by beebe
This might be more of a trig question, so if it is would someone kindly move it for me?

So I'm finding the second derivative of $ln[sec(x)+tan(x)]$. My first derivative is $\frac{sec(x)tan(x)+sec^2(x)}{sec(x)+tan(x)}$.

My question is: can I simplify this fraction to make taking the second derivative less tedious? I'm capable of doing it without simplifying first, but I don't want to get a super long and ugly answer.
Or you could multiply top and bottom by $\displaystyle \sec{x} - \tan{x}$, since you should know that $\displaystyle \sec^2{x} - \tan^2{x} \equiv 1$...

4. ## Re: Simplifying a denominator before taking derivative

Thanks guys.

Originally Posted by TKHunny
Have you considered factoring out sec(x) from the numerator?
I feel kinda dumb for missing this one.

5. ## Re: Simplifying a denominator before taking derivative

Originally Posted by beebe
This might be more of a trig question, so if it is would someone kindly move it for me?

So I'm finding the second derivative of $ln[sec(x)+tan(x)]$. My first derivative is $\frac{sec(x)tan(x)+sec^2(x)}{sec(x)+tan(x)}$.

My question is: can I simplify this fraction to make taking the second derivative less tedious? I'm capable of doing it without simplifying first, but I don't want to get a super long and ugly answer.
yes.