If x^p y^q = (x + y)^ p + q , then d^2 y d x^2 = - Vidyadip

MCQ

If ${x^p}{y^q} = {(x + y)^{p + q}}$, then $\frac{{{d^2}y}}{{d{x^2}}} = $

✓
$0$
B
$1$
C
$2$
D
None of these

✓

Answer

Correct option: A.

$0$

a
(a) Taking $\log $ and differentiating, we get $x\frac{{dy}}{{dx}} - y = 0$
Differentiating again, we get $\frac{{{d^2}y}}{{d{x^2}}} = 0$.

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