Question

Question #7d154

Answer 1

`y' = 0`

Explanation:

We know that `sin^2x + cos^2x = 1`, therefore, `y = 1`. The derivative of a constant is `0`, thus `y' = 0`.

Hopefully this helps!

Answer 2

0

Explanation:

If `y = sin^2x + cos^2x`, then y = 1 (a constant).

And the derivative would be zero.

Or, you can have a 'senior moment' like I did, and forget this, for a moment, and go through the motions of actually deriving the function. It's the SUM of terms `sin^2x` and `cos^2x`, so you can take the derivative of each of the terms and then combine them:

`d/dx sin^2x = 2sinxcosx` (hint: use the CHAIN RULE)

`d/dx cos^2x = -2cosxsinx`

So, you'd have `d/dx (sin^2x + cos^2x) =`

`2sinxcosx -2cosxsinx =`

0 (once again).

GOOD LUCK