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

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 16 May 2024 Shift 2 Paper

Question : 119 of 150

Marks: +1, -0

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

\frac{d y}{d x}=

Solution:

Concept:

Use standard inverse trigonometric identities to simplify the expression before differentiating.

Explanation:

Rewrite the first term using the formula for

\tan^{-1}a + \tan^{-1}b

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

Rewrite the second term using

4x = 5x - x

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

Add both simplified parts:

y = \tan^{-1}\frac{2}{3} + \tan^{-1}x + \tan^{-1}5x - \tan^{-1}x = \tan^{-1}\frac{2}{3} + \tan^{-1}5x.

Differentiate with respect to

x

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

Answer:

Option B:

\frac{5}{1+25x^2}

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