Question

(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 for the function \(g(x) = \sin^2(x) + \cos^2(x)\) is 0. For the expressions \(f(x)=\sec ^{2} x\) and \(g(x)=\tan ^{2} x\), it was found that their derivatives are equivalent as \(f^{\prime}(x) = g^{\prime}(x) = 2\sec^2(x)\tan(x)\).

Step-by-step solution

Differentiate g(x) for Part (a)

To differentiate \(g(x)=\sin ^{2} x+\cos ^{2} x\), we need to differentiate each term separately. The derivative of \(\sin ^{2} x\) is \(2\sin(x)\cos(x)\) using the chain rule. Similarly, the derivative of \(\cos ^{2} x\) is \(-2\sin(x)\cos(x)\), again applying the chain rule. Summing them up, we get \(2\sin(x)\cos(x)-2\sin(x)\cos(x) = 0\).

Utilize Trigonometric Identity for Part (a)

The task asked to find the derivative in two ways. This was the first way. Alternatively, by using the trigonometric identity \(\sin ^{2} x+\cos ^{2} x = 1\), we can see that \(g(x) = 1\), a constant. Hence its derivative is \(0\). As such, both ways results in the same derivative.

Differentiate f(x) for Part (b)

For \(f(x)=\sec ^{2} x\), its derivative is \(f^{\prime}(x) = 2\sec(x)\sec(x)\tan(x) = 2\sec^2(x)\tan(x)\).

Differentiate g(x) for Part (b)

For \(g(x)=\tan ^{2} x\), its derivative \( g^{\prime}(x) = 2\tan(x)\sec^2(x) = 2\sec^2(x)\tan(x) \).

Evaluate Derived Expressions for Part (b)

The derivatives \(f^{\prime}(x)\) and \(g^{\prime}(x)\) in Step 3 and Step 4 are indeed equal. Hence, we have shown that \(f^{\prime}(x) = g^{\prime}(x)\).