# Question #7d154

Dec 14, 2017

$y ' = 0$

#### Explanation:

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

Hopefully this helps!

Dec 14, 2017

0

#### Explanation:

If $y = {\sin}^{2} x + {\cos}^{2} x$, 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 ${\sin}^{2} x$ and ${\cos}^{2} x$, so you can take the derivative of each of the terms and then combine them:

$\frac{d}{\mathrm{dx}} {\sin}^{2} x = 2 \sin x \cos x$ (hint: use the CHAIN RULE)

$\frac{d}{\mathrm{dx}} {\cos}^{2} x = - 2 \cos x \sin x$

So, you'd have $\frac{d}{\mathrm{dx}} \left({\sin}^{2} x + {\cos}^{2} x\right) =$

$2 \sin x \cos x - 2 \cos x \sin x =$

0 (once again).

GOOD LUCK