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Exercise 10.5 · Q15

Q.If y=(1−x)2x2y = \dfrac{(1-x)^2}{x^2}, then dydx\dfrac{dy}{dx} is

(1) 2x2+2x3\dfrac{2}{x^2}+\dfrac{2}{x^3}
(2) −2x2+2x3-\dfrac{2}{x^2}+\dfrac{2}{x^3}
(3) −2x2−2x3-\dfrac{2}{x^2}-\dfrac{2}{x^3}
(4) −2x3+2x2-\dfrac{2}{x^3}+\dfrac{2}{x^2}
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Step 1. Expand the numerator: (1−x)2=1−2x+x2(1-x)^2=1-2x+x^2, so

y=1−2x+x2x2=1x2−2x+1=x−2−2x−1+1y=\frac{1-2x+x^2}{x^2}=\frac{1}{x^2}-\frac{2}{x}+1=x^{-2}-2x^{-1}+1

Step 2. Differentiate term by term:

dydx=−2x−3−2(−1)x−2+0=−2x3+2x2\frac{dy}{dx}=-2x^{-3}-2(-1)x^{-2}+0=-\frac{2}{x^3}+\frac{2}{x^2}

Step 3. (Cross-check via quotient rule.) With u=(1−x)2u=(1-x)^2, v=x2v=x^2: u′=−2(1−x)u'=-2(1-x), v′=2xv'=2x. …

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