Question

Using Trigonometric Functions (a) Find the derivative of the function \(g(x)=\sin ^{2} x+\cos ^{2} x\) in two ways. (b) For \(f(x)=\sec ^{2} x\) and \(g(x)=\tan ^{2} x,\) show that \(f^{\prime}(x)=g^{\prime}(x)\)

Short answer

The derivative of \(g(x) = \sin^2 x + \cos^2 x\) is 0. The derivatives of \(f(x) = \sec^2 x\) and \(g(x) = \tan^2 x\) are both equal to \(2 \sec^2x \tan x\).

Step-by-step solution

First derivative for part (a)

For the function \(g(x) = \sin^2 x + \cos^2 x\), note that this sum \(sin^2x + cos^2x\) is a well-known identity that equals to 1. Therefore, \(g(x) = 1\), and the derivative of a constant is 0. Hence, \(g'(x) = 0\).

Second derivative for part (a)

For a deeper understanding, we can differentiate the given function using the chain rule too,\[\frac{d}{dx}(\sin^2x) = 2\sin x \cos x\]\[\frac{d}{dx}(\cos^2x) = -2\sin x \cos x\]So \(g'(x) = 2\sin x \cos x - 2\sin x \cos x = 0\]

Derivative for \(f(x)\) in part (b)

For the function \(f(x) = \sec^2 x\), the derivative is calculated as \(f'(x) = 2 \sec{x} \cdot \sec{x} \tan{x}\)

Derivative for \(g(x)\) in part (b)

For the function \(g(x) = \tan^2 x\), the derivative is calculated as \(g'(x) = 2 \tan{x} \cdot \sec^2{x}\)

Verify the equality of derivatives for part (b)

From step 3 and 4, it can be seen that both \(f'(x)\) and \(g'(x)\) simplify to \(2 \sec^2{x} \tan{x}\) therefore we confirm that \(f'(x)=g'(x)\)