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EXERCISE 1.3 · Q142

Q.If log⁡10x3−y3x3+y3=2\log_{10}\dfrac{x^3-y^3}{x^3+y^3}=2, show dydx=−99x2101y2\dfrac{dy}{dx}=-\dfrac{99x^2}{101y^2}.

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Given log⁡10x3−y3x3+y3=2\log_{10}\dfrac{x^3-y^3}{x^3+y^3}=2.

Step 1 — undo the log: log⁡10(⋅)=2  ⟹  (⋅)=102=100\log_{10}(\cdot)=2 \implies (\cdot)=10^2=100.

x3−y3x3+y3=100  ⟹  x3−y3=100(x3+y3)\frac{x^3-y^3}{x^3+y^3}=100 \implies x^3-y^3=100(x^3+y^3)

Step 2 — differentiate this relation implicitly:

3x2−3y2dydx=100(3x2+3y2dydx)3x^2-3y^2\frac{dy}{dx}=100\left(3x^2+3y^2\frac{dy}{dx}\right)

3x2−3y2dydx=300x2+300y2dydx3x^2-3y^2\frac{dy}{dx}=300x^2+300y^2\frac{dy}{dx}

Step 3 — collect dy/dxdy/dx terms:

3x2−300x2=300y2dydx+3y2dydx3x^2-300x^2=300y^2\frac{dy}{dx}+3y^2\frac{dy}{dx} …

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