Method of DifferentiationmediumFree

Method of Differentiation — JEE Maths practice question

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

Question
y=2x2(x−2)(x−3)(x−4)+3x(x−3)(x−4)+xx−4y = \dfrac{2x^2}{(x-2)(x-3)(x-4)} + \dfrac{3x}{(x-3)(x-4)} + \dfrac{x}{x-4}. If xydydx=a2−x+b3−x+c4−x\dfrac{x}{y}\dfrac{dy}{dx} = \dfrac{a}{2-x}+\dfrac{b}{3-x}+\dfrac{c}{4-x}, then a+b+ca+b+c is.
Solution
Answer: 9
Step 1: Simplify yy Combining over the common denominator (x−2)(x−3)(x−4)(x-2)(x-3)(x-4):
y=2x2+3x(x−2)+x(x−2)(x−3)(x−2)(x−3)(x−4)=x3(x−2)(x−3)(x−4)y = \frac{2x^2 + 3x(x-2) + x(x-2)(x-3)}{(x-2)(x-3)(x-4)} = \frac{x^3}{(x-2)(x-3)(x-4)}
 Step 2: Take logarithms and differentiate
log⁡y=3log⁡x−log⁡(x−2)−log⁡(x−3)−log⁡(x−4)\log y = 3\log x - \log(x-2) - \log(x-3) - \log(x-4)
xydydx=3−xx−2−xx−3−xx−4\frac{x}{y}\frac{dy}{dx} = 3 - \frac{x}{x-2} - \frac{x}{x-3} - \frac{x}{x-4}
 Step 3: Convert to the required form Using xx−k=1−kk−x\dfrac{x}{x-k} = 1 - \dfrac{k}{k-x}:
xydydx=22−x+33−x+44−x\frac{x}{y}\frac{dy}{dx} = \frac{2}{2-x} + \frac{3}{3-x} + \frac{4}{4-x}
 So a=2a = 2, b=3b = 3, c=4c = 4, and a+b+c=9a+b+c = 9. Answer: 9
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