If y=≤ft(a x+b/c x+d), then 2 d y/d x · {d³ y}{d x³} is equal to

Correct Answer is : 3(d2ydx2)23≤ft({d² y}{d x²})²3(dx2d2y)2

Correct Answer is : 3(d2ydx2)23≤ft({d² y}{d x²})²3(dx2d2y)2

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MHTCET 15 Apr 2026 Shift 1 Paper

Question : 122 of 150

Marks: +1, -0

y=\left(\frac{a x+b}{c x+d}\right)

2 \frac{d y}{d x} \cdot \frac{d^{3} y}{d x^{3}}

is equal to

Solution:

Concept:

We differentiate a rational function repeatedly and compare the product

2\frac{dy}{dx}\cdot\frac{d^{3}y}{dx^{3}}

with

\left(\frac{d^{2}y}{dx^{2}}\right)^{2}

Explanation:

Given

y=\frac{ax+b}{cx+d}

, differentiate using the quotient rule:

\frac{dy}{dx}=\frac{a(cx+d)-c(ax+b)}{(cx+d)^{2}}=\frac{ad-bc}{(cx+d)^{2}}

Let

ad-bc=k

, so

\frac{dy}{dx}=k(cx+d)^{-2}

Differentiate again:

\frac{d^{2}y}{dx^{2}}=-2ck(cx+d)^{-3}

Differentiate once more:

\frac{d^{3}y}{dx^{3}}=6c^{2}k(cx+d)^{-4}

Now compute the required product:

2\frac{dy}{dx}\cdot\frac{d^{3}y}{dx^{3}}=2\left[k(cx+d)^{-2}\right]\left[6c^{2}k(cx+d)^{-4}\right]

=12c^{2}k^{2}(cx+d)^{-6}

Next, square the second derivative:

\left(\frac{d^{2}y}{dx^{2}}\right)^{2}=\left[-2ck(cx+d)^{-3}\right]^{2}=4c^{2}k^{2}(cx+d)^{-6}

Therefore,

3\left(\frac{d^{2}y}{dx^{2}}\right)^{2}=12c^{2}k^{2}(cx+d)^{-6}

Hence,

2\frac{dy}{dx}\cdot\frac{d^{3}y}{dx^{3}}=3\left(\frac{d^{2}y}{dx^{2}}\right)^{2}

Answer:

Option C:

3\left(\frac{d^{2}y}{dx^{2}}\right)^{2}

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