If log _{10}left(frac{x^3-y^3}{x^3+y^3}right) | vedclass

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(D) Given,$\log _{10}\left(\frac{x^3-y^3}{x^3+y^3}\right)=2$ $\Rightarrow \frac{x^3-y^3}{x^3+y^3} = 10^2 = 100$ $\Rightarrow x^3 - y^3 = 100x^3 + 100y

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$\frac{y}{x}$

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(D) Given,$\log _{10}\left(\frac{x^3-y^3}{x^3+y^3}\right)=2$ $\Rightarrow \frac{x^3-y^3}{x^3+y^3} = 10^2 = 100$ $\Rightarrow x^3 - y^3 = 100x^3 + 100y^3$ $\Rightarrow -99x^3 = 101y^3$ $\Rightarrow y^3 = -\frac{99}{101}x^3$ Differentiating both sides with respect to $x$: $3y^2 \frac{dy}{dx} = -\frac{99}{101} \cdot 3x^2$ $\frac{dy}{dx} = -\frac{99}{101} \cdot \frac{x^2}{y^2}$ Since $y^3 = -\frac{99}{101}x^3$,we have $-\frac{99}{101} = \frac{y^3}{x^3}$ Substituting this into the derivative: $\frac{dy}{dx} = \frac{y^3}{x^3} \cdot \frac{x^2}{y^2} = \frac{y}{x}$

If $\log _{10}\left(\frac{x^3-y^3}{x^3+y^3}\right)=2$,then $\frac{d y}{d x}=$

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