# What is the derivative of sin^2x+cos^2x?

May 16, 2017

We have:

$f \left(x\right) = {\sin}^{2} x + {\cos}^{2} x$

We can differentiate using the chain rule to get:

$f ' \left(x\right) = 2 \sin x \frac{d}{\mathrm{dx}} \left(\sin x\right) + 2 \cos x \frac{d}{\mathrm{dx}} \left(\cos x\right)$
$\text{ } = 2 \sin x \cos x + 2 \cos x \left(- \sin x\right)$
$\text{ } = 2 \sin x \cos x - 2 \sin x \cos x$
$\text{ } = 0 \setminus \setminus \setminus$ QED

However, it really should be clear that the above calculation was completely unnecessary and the result is obvious.

A fundamental trigonometric identity is:

${\sin}^{2} A + {\cos}^{2} A \equiv 1 \setminus \setminus \setminus \forall A \in \mathbb{R}$

Hence we have:

$f \left(x\right) = 1 \setminus \setminus \setminus \forall x \in \mathbb{R}$

ie, $f \left(x\right)$ is a constant function, and the derivative of constant is zero.