What happens when we differentiate $\sin^4x + 2 \cos^2x - \cos^4x$?

What happens when we differentiate $\sin^4x + 2 \cos^2x - \cos^4x$?

Question

Verify that the differential coefficient with respect to

$x$

of

$$\begin{equation*} \sin^4 x + 2 \cos^2 x - \cos^4 x \end{equation*}$$

is equal to $0$ for all values of $x$. What deduction can be made from this result?

[By ‘the differential coefficient’, the question means ‘the derivative’.]

Solution

By the

chain rule

, we have that

$$\begin{equation*} \frac{d}{dx} \sin^4 x = 4 \sin^3 x \frac{d}{dx} \sin x = 4 \sin^3 x \cos x \end{equation*}$$

and

$$\begin{equation*} \frac{d}{dx} \cos^2 x = 2 \cos x \frac{d}{dx} \cos x = -2 \cos x \sin x \end{equation*}$$

and

$$\begin{equation*} \frac{d}{dx} \cos^4 x = 4 \cos^3 x \frac{d}{dx} \cos x = - 4 \cos^3 x \sin x. \end{equation*}$$

Therefore

$$\begin{align*} \frac{d}{dx} \left( \sin^4 x + 2 \cos^2 x - \cos^4 x \right) &= \frac{d}{dx} \sin^4 x + 2 \frac{d}{dx} \cos^2 x - \frac{d}{dx} \cos^4 x \\ &= 4 \sin^3 x \cos x - 4 \cos x \sin x + 4 \cos^3 x \sin x \\ &= 4 \sin x \cos x \left( \sin^2 x + \cos^2 x - 1 \right) \\ &= 0, \end{align*}$$

as required.

We can use $c$ for $\cos x$ and $s$ for $\sin x$ in the above to simplify things (not forgetting that $s^2 + c^2 = 1$).

We can conclude from this that

$$\begin{equation*} \sin^4 x + 2 \cos^2 x - \cos^4 x \end{equation*}$$

is constant for every value of $x$. By substituting $x = 0$ into the above, this constant must equal $0 + 2 - 1 = 1$.

Alternatively, we can say

$$\begin{align*} \sin^4 x + 2 \cos^2 x - \cos^4 x &= s^4-c^4+2c^2\\ &= (s^2+c^2)(s^2-c^2)+2c^2 \\ &= s^2 -c^2 + 2c^2\\ &= s^2 + c^2 =1. \end{align*}$$