Question #7d154

2 Answers
Dec 14, 2017

#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!

Dec 14, 2017

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