Differentiation Question 371
Question: If $ y=x\log ( \frac{x}{a+bx} ) $ , then $ x^{3}\frac{d^{2}y}{dx^{2}}= $
[WB JEE 1991; Roorkee 1976]
Options:
A) $ x\frac{dy}{dx}-y $
B) $ {{( x\frac{dy}{dx}-y )}^{2}} $
C) $ y\frac{dy}{dx}-x $
D) $ {{( y\frac{dy}{dx}-x )}^{2}} $
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Answer:
Correct Answer: B
Solution:
From the given relation $ \frac{y}{x}=\log x-\log (a+bx) $
Differentiating we get $ \frac{( x\frac{dy}{dx}-y )}{x^{2}}=\frac{1}{x}-\frac{1}{a+bx}b=\frac{a}{x(a+bx)} $
$ \therefore x\frac{dy}{dx}-y=\frac{ax}{a+bx} $
…..(i) Differentiating again w.r.t. x, we get $ x\frac{d^{2}y}{dx^{2}}+\frac{dy}{dx}-\frac{dy}{dx}=\frac{(a+bx)a-ax.b}{{{(a+bx)}^{2}}} $
Therefore $ x\frac{d^{2}y}{dx^{2}}=\frac{a^{2}}{{{(a+bx)}^{2}}} $
Therefore $ x^{3}\frac{d^{2}y}{dx^{2}}=\frac{a^{2}x^{2}}{{{(a+bx)}^{2}}}={{( x\frac{dy}{dx}-y )}^{2}} $ , [by (i)].
BETA
