Method of DifferentiationmediumFree

Method of Differentiation — JEE Maths practice question

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

Question
If y=xsin⁡(log⁡x)+xlog⁡xy = x\sin(\log x) + x\log x, then x2d2ydx2−xdydx+2yx^2\dfrac{d^2 y}{dx^2} - x\dfrac{dy}{dx} + 2y is:
Aln⁡x\ln x
Bxlog⁡xx\log xcorrect
C−xln⁡x-x\ln x
Dln⁡xx\dfrac{\ln x}{x}
Solution
Step 1: First derivative
y1=sin⁡(log⁡x)+cos⁡(log⁡x)+log⁡x+1y_1 = \sin(\log x) + \cos(\log x) + \log x + 1
 Step 2: Second derivative
y2=cos⁡(log⁡x)x−sin⁡(log⁡x)x+1xy_2 = \frac{\cos(\log x)}{x} - \frac{\sin(\log x)}{x} + \frac{1}{x}
 Step 3: Substitute into the expression
x2y2=xcos⁡(log⁡x)−xsin⁡(log⁡x)+xx^2 y_2 = x\cos(\log x) - x\sin(\log x) + x
xy1=xsin⁡(log⁡x)+xcos⁡(log⁡x)+xlog⁡x+xx y_1 = x\sin(\log x) + x\cos(\log x) + x\log x + x
x2y2−xy1+2y=xlog⁡xx^2 y_2 - x y_1 + 2y = x\log x
 Answer: (2)
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