Method of DifferentiationhardFree

Method of Differentiation — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question

y=\dfrac{\alpha x+\beta}{\gamma x+\delta}

, then

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

7\!\left(\dfrac{d^{2}y}{dx^{2}}\right)^{2}

5\!\left(\dfrac{d^{2}y}{dx^{2}}\right)^{2}

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

correct

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

Solution

Step 1: Let

D=\alpha\delta-\beta\gamma

. Compute the first three derivatives:

y_{1}=\dfrac{D}{(\gamma x+\delta)^{2}}

y_{2}=\dfrac{-2D\gamma}{(\gamma x+\delta)^{3}}

y_{3}=\dfrac{6D\gamma^{2}}{(\gamma x+\delta)^{4}}

Step 2: Compute

2y_{1}y_{3}

2y_{1}y_{3}=2\cdot\dfrac{D}{(\gamma x+\delta)^{2}}\cdot\dfrac{6D\gamma^{2}}{(\gamma x+\delta)^{4}}=\dfrac{12D^{2}\gamma^{2}}{(\gamma x+\delta)^{6}}

Step 3: Compute

3(y_{2})^{2}

3y_{2}^{2}=3\cdot\left(\dfrac{-2D\gamma}{(\gamma x+\delta)^{3}}\right)^{2}=3\cdot\dfrac{4D^{2}\gamma^{2}}{(\gamma x+\delta)^{6}}=\dfrac{12D^{2}\gamma^{2}}{(\gamma x+\delta)^{6}}

Step 4: Therefore:

2y_{1}y_{3}=3y_{2}^{2}\;\Rightarrow\;2\dfrac{dy}{dx}\cdot\dfrac{d^{3}y}{dx^{3}}=3\left(\dfrac{d^{2}y}{dx^{2}}\right)^{2}

Correct answer: (3)

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