Let's do some calculus! (12)

Calculus Level 4

Given that y = a x 2 ( x − a ) ( x − b ) ( x − c ) + b x ( x − b ) ( x − c ) + c x − c + 1 y= \dfrac{ax^{2}}{(x-a)(x-b)(x-c)} + \dfrac{bx}{(x-b)(x-c)} + \dfrac{c}{x-c} + 1

Find y ′ y \dfrac{y'}{y} .

Notation: y ′ y' denotes the first derivative of y y .

For more problems on calculus, click here .
1 x ( a a − x + b b − x + c c − x ) \dfrac{1}{x} \left( \dfrac{a}{a-x} + \dfrac{b}{b-x} + \dfrac{c}{c-x} \right) 1 a b c ⋅ 1 x ( a a − x + b b − x + c c − x ) \dfrac{1}{abc}\cdot\dfrac{1}{x} \left( \dfrac{a}{a-x} + \dfrac{b}{b-x} + \dfrac{c}{c-x} \right) 1 x ( a a + x + b b + x + c c + x ) \dfrac{1}{x} \left( \dfrac{a}{a+x} + \dfrac{b}{b+x} + \dfrac{c}{c+x} \right) 1 a b c ⋅ 1 x ( a a + x + b b + x + c c + x ) \dfrac{1}{abc}\cdot\dfrac{1}{x} \left( \dfrac{a}{a+x} + \dfrac{b}{b+x} + \dfrac{c}{c+x} \right)

Tapas Mazumdar by Tapas Mazumdar , 20, India

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1 solution

Simple simlification shows y = x 3 ( x − a ) ( x − b ) ( x − c ) \displaystyle y=\frac{x^3}{(x-a)(x-b)(x-c)} , now differentiating we have y ′ y = 3 x − 1 x − a − 1 x − b − 1 x − c = 1 x ( a a − x + b b − x + c c − x ) \displaystyle \frac{y'}{y}=\frac{3}{x}-\frac{1}{x-a}-\frac{1}{x-b}-\frac{1}{x-c}=\frac{1}{x}(\frac{a}{a-x}+\frac{b}{b-x}+\frac{c}{c-x})

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