The derivative of ≤ft( ≤ft(x+3/x)) is

Correct Answer is : −3x(x+3)cos⁡(log⁡(x+3x))-3/x(x+3) ≤ft( ≤ft(x+3/x))x(x+3)−3cos(log(xx+3))

Correct Answer is : −3x(x+3)cos⁡(log⁡(x+3x))-3/x(x+3) ≤ft( ≤ft(x+3/x))x(x+3)−3cos(log(xx+3))

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MHTCET 15 Apr 2026 Shift 2 Paper

Question : 122 of 150

Marks: +1, -0

\sin \left(\log \left(\frac{x+3}{x}\right)\right)

Solution:

Concept:

Use the chain rule and the derivative of a logarithmic function.

Explanation:

Let

y = \sin\left(\log\left(\frac{x+3}{x}\right)\right)

Set

u = \log\left(\frac{x+3}{x}\right)

Using

\log\left(\frac{a}{b}\right) = \log a - \log b

, we get

u = \log(x+3) - \log x

Differentiate

u

with respect to

x

\frac{du}{dx} = \frac{1}{x+3} - \frac{1}{x}

Taking LCM,

\frac{du}{dx} = \frac{x - (x+3)}{x(x+3)} = \frac{-3}{x(x+3)}

Now apply the chain rule:

\frac{dy}{dx} = \cos\left(\log\left(\frac{x+3}{x}\right)\right) \cdot \frac{du}{dx}

Substitute

\frac{du}{dx}

\frac{dy}{dx} = \frac{-3}{x(x+3)} \cos\left(\log\left(\frac{x+3}{x}\right)\right)

Answer:

The correct option is Option C:

\frac{-3}{x(x+3)} \cos\left(\log\left(\frac{x+3}{x}\right)\right)

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