Question

Assertion (A): \[ \frac{d}{dx}\left(\frac{x^2\sin x}{\log x}\right) = \frac{x^2\sin x}{\log x} \left(\cot x+\frac{2}{x}-\frac{1}{x\log x}\right) \] Reason (R): \[ \frac{d}{dx}\left(\frac{uv}{w}\right) = \frac{uv}{w} \left[ \frac{u'}{u}+\frac{v'}{v}-\frac{w'}{w} \right] \]

• A. A is true, R is true and R is correct explanation of A
• B. A is true, R is true and R is not correct explanation of A
• C. A is true, R is not correct
• D. A is not correct, R is correct

Hint:

For functions involving products, quotients, and powers, logarithmic differentiation often makes the calculation easier and faster.

Answer 1

The Correct Option is A

Step 1: Verify the Reason (R) using logarithmic differentiation.
Let $y = uv/w$. Taking natural log: $\ln y = \ln u + \ln v - \ln w$. Differentiating: $\frac{1}{y}\frac{dy}{dx} = \frac{u'}{u} + \frac{v'}{v} - \frac{w'}{w}$, so $\frac{dy}{dx} = \frac{uv}{w}\left(\frac{u'}{u} + \frac{v'}{v} - \frac{w'}{w}\right)$. Reason (R) is TRUE.
Step 2: Apply the formula from Reason (R) to verify Assertion (A).
Let $y = \frac{x^2 \sin x}{\log x}$. Here $u = x^2$, $v = \sin x$, $w = \log x$.
Step 3: Compute the individual logarithmic derivatives.
$\frac{u'}{u} = \frac{2x}{x^2} = \frac{2}{x}$. $\frac{v'}{v} = \frac{\cos x}{\sin x} = \cot x$. $\frac{w'}{w} = \frac{1/x}{\log x} = \frac{1}{x \log x}$.
Step 4: Substitute into the formula from Reason (R).
$\frac{dy}{dx} = \frac{x^2 \sin x}{\log x} \left(\frac{2}{x} + \cot x - \frac{1}{x \log x}\right) = \frac{x^2 \sin x}{\log x}\left(\cot x + \frac{2}{x} - \frac{1}{x \log x}\right)$. This matches exactly what Assertion (A) states, so (A) is TRUE.
Step 5: Check if R correctly explains A.
The formula in (R) is the precise tool used to derive (A). So (R) is indeed the correct explanation of (A).
Step 6: Select the correct option.
Since both (A) and (R) are true and (R) is the correct explanation of (A), the answer is option 1. \[\boxed{\text{A is true, R is true and R is the correct explanation of A}}\]