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: Identify g:
Differentiating $\ln y = \sin x\ln(x^2+1)$ gives the two terms $\frac{2x\sin x}{x^2+1}$ and $\cos x\ln(x^2+1)$, so $g(x) = x^2+1$.

Step 2: Look at the function:
$h(x) = (x^2+1)^{-1}$. As $x$ grows beyond 0, $x^2+1$ grows, so its reciprocal falls.

Step 3: Derivative test:
$h'(x) = -2x(x^2+1)^{-2} < 0$ for all $x > 0$, so $h$ is strictly decreasing there.

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