JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Analyzing +4 | -1

y = x\log \left( {\frac{x}{{a + bx}}} \right),

then

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

is equal to

View answer and explanation

Correct option: B

{\left( {x\frac{{dy}}{{dx}} - y} \right)^2}

The correct answer is B (

{\left( {x\frac{{dy}}{{dx}} - y} \right)^2}

) because it is the result of differentiating the given function and substituting to find the expression for

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

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x

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