Verify that the differential coefficient with respect to \(x\) of
\[\begin{equation*}
\sin^4 x + 2 \cos^2 x - \cos^4 x
\end{equation*}\]
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
\[\begin{align*}
\frac{d}{dx} \left( \sin^4 x + 2 \cos^2 x - \cos^4 x \right) &= \frac{d}{dx} \sin^4 x + 2 \frac{d}{dx} \cos^2 x - \frac{d}{dx} \cos^4 x \\
&= 4 \sin^3 x \cos x - 4 \cos x \sin x + 4 \cos^3 x \sin x \\
&= 4 \sin x \cos x \left( \sin^2 x + \cos^2 x - 1 \right) \\
&= 0,
\end{align*}\]
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’.]
as required.
We can use \(c\) for \(\cos x\) and \(s\) for \(\sin x\) in the above to simplify things (not forgetting that \(s^2 + c^2 = 1\)).
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
\[\begin{align*}
\sin^4 x + 2 \cos^2 x - \cos^4 x &= s^4-c^4+2c^2\\
&= (s^2+c^2)(s^2-c^2)+2c^2 \\
&= s^2 -c^2 + 2c^2\\
&= s^2 + c^2 =1.
\end{align*}\]
