Question

If \(y = (x^2+1)^{sinx}\) for \(x > 0\) such that \(\frac{dy}{dx} = y[\frac{2xsinx}{g(x)}+cosx\cdot log[g(x)]]\), then the function \(\frac{1}{g(x)}\) is...

• A. increasing
• B. strictly increasing.
• C. decreasing
• D. strictly decreasing

Hint:

Differentiate y = (x squared + 1)^{sin x} logarithmically to identify g(x).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Take logarithms: \(\log y = \sin x \cdot \log(x^2+1)\), then differentiate to compare with the given form.

Step 2: Differentiate:
\[ \frac{1}{y}\frac{dy}{dx} = \cos x\log(x^2+1) + \sin x \cdot \frac{2x}{x^2+1} \]
So \(\frac{dy}{dx} = y\left[\frac{2x\sin x}{x^2+1} + \cos x\log(x^2+1)\right]\).
Comparing with the given form, \(g(x) = x^2 + 1\).

Step 3: Monotonicity of 1/g:
\(\frac{1}{g(x)} = \frac{1}{x^2+1}\). Its derivative is \(-\frac{2x}{(x^2+1)^2}\), which is negative for \(x > 0\).
So \(\frac1{g(x)}\) is strictly decreasing for \(x > 0\).

Step 4: Options:
Option (D), strictly decreasing, is the precise description that holds for \(x > 0\). The increasing options are false because the derivative is negative.

Final Answer:
\(1/g(x)\) is strictly decreasing, option (D). \[ \boxed{\text{Strictly decreasing (D)}} \]