Why is d/dx 4 * (sec x)^2 = 8 * (sec x)^2 * tan x

Question

anonymous · Answer

While (sin x)^2 = 2 - sin x * cos x 

More specifically: I am wondering about why sec remains squared in the first equation but sin does not in the second.

kainui · Answer

It's not really remaining squared in the first one. When you do the chain rule you go from (secx)^2 to secx however that's only the outside part. The derivative of the inside part is the derivative of secx itself, which is secx*tanx. So when we multiply by this, we have (secx)^2 again.

kainui · Answer

To be maybe a little more clear, $$\Large \frac{d}{dx}(4( \sec x)^2)=4*(2 \sec x)* \sec x \tan x$$

anonymous · Answer

Thanks you very much!

kainui · Answer

You're welcome, I hope that clears it up. You're right on the edge of knowing how calculus works so that you can start to understand how to use it like you would adding.