If y= ^{-1}≤ft({4 x}{1+5 x²})+ ^{-1}≤ft(3-2 x/2+3 x), then d y/d x is equal to

Correct Answer is : 51+25x2{5}{1+25 x²}1+25x25

Correct Answer is : 51+25x2{5}{1+25 x²}1+25x25

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MHTCET 25 Apr 2025 Shift 1 Paper

Question : 112 of 150

Marks: +1, -0

y=\tan ^{-1}\left(\frac{4 x}{1+5 x^{2}}\right)+\cot ^{-1}\left(\frac{3-2 x}{2+3 x}\right)

\frac{d y}{d x}

is equal to

Solution:

Concept:

Use inverse tangent addition and subtraction formulas to simplify

y

before differentiating.

Explanation:

Rewrite

\cot^{-1}\left(\frac{3-2x}{2+3x}\right)

\tan^{-1}\left(\frac{2+3x}{3-2x}\right)

Now simplify each term using standard identities.

First term:

\frac{4x}{1+5x^2} = \frac{5x-x}{1+(5x)(x)}

So,

\tan^{-1}\left(\frac{4x}{1+5x^2}\right) = \tan^{-1}(5x) - \tan^{-1}x

Second term:

\frac{2+3x}{3-2x} = \frac{\frac{2}{3}+x}{1-\frac{2}{3}x}

So,

\tan^{-1}\left(\frac{2+3x}{3-2x}\right) = \tan^{-1}\left(\frac{2}{3}\right) + \tan^{-1}x

Adding both results,

y = \tan^{-1}(5x) + \tan^{-1}\left(\frac{2}{3}\right)

Differentiate with respect to

x

\frac{dy}{dx} = \frac{1}{1+(5x)^2} \cdot 5 = \frac{5}{1+25x^2}

Answer:

\frac{dy}{dx} = \frac{5}{1+25x^2}

Hence, the correct option is A.

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