JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Hard

1 of 1

Analyzing +4 | -1

y = \left( {\frac{{ax + b}}{{cx + d}}} \right),

then

2\frac{{dy}}{{dx}}.\frac{{{d^3}y}}{{d{x^3}}}

is equal to

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because after differentiating the given function three times, the expression

2\frac{{dy}}{{dx}}.\frac{{{d^3}y}}{{d{x^3}}}

simplifies to

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

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