Question

If $\log_{10} \left(\frac{x^{3} - y^{3} }{x^{3} + y^3} \right) = 2$ then $ \frac{dy}{dx} = $

• A. $\frac{x}{y}$
• B. $- \frac{y}{x}$
• C. $ - \frac{x}{y}$
• D. $ \frac{y}{x}$

Answer 1

The Correct Option is 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}=100\left(x^{3}+y^{3}\right) $
$ \Rightarrow 101 y^{3}=-99 x^{3}$
On differentiating both sides w.r.t. $x$ we get
$101 \times 3 y^{2} \frac{d y}{d x}=-99 \cdot\left(3 x^{2}\right)$
$\Rightarrow 101 y^{2} \frac{d y}{d x}=-99 x^{2}$
On multiplying by $x$ both sides, we get
$\Rightarrow 101 x y^{2} \frac{d y}{d x}=-99 x^{3}$
$\Rightarrow \frac{d y}{d x}=\frac{-99 x^{3}}{101 x y^{2}}$
$\Rightarrow \frac{d y}{d x}=\frac{101 y^{3}}{101 x y^{2}} \left[\because-99 x^{3}=101 y^{3}\right]$
$\Rightarrow \frac{d y}{d x}=\frac{y}{x}$

Answer 2

Given:

\(\log_{10} \left(\frac{x^{3} - y^{3}}{x^{3} + y^{3}}\right) = 2,\)

we start by rewriting the logarithmic equation in its exponential form:
\(\frac{x^{3} - y^{3}}{x^{3} + y^{3}} = 10^{2} = 100\)

\(x^{3} - y^{3} = 100(x^{3} + y^{3})\)

Rearrange and combine like terms:
\(x^{3} - y^{3} = 100x^{3} + 100y^{3},\)
\(x^{3} - 100x^{3} - y^{3} - 100y^{3} = 0,\)
\(-99x^{3} - 101y^{3} = 0,\)
\(99x^{3} = -101y^{3},\)

\(\left(\frac{x}{y}\right)^{3} = -\frac{101}{99},\)

\(99x^{3} + 101y^{3} = 0\)

Taking the derivative with respect to x:
\(99 \cdot 3x^{2} + 101 \cdot 3y^{2} \frac{dy}{dx} = 0,\)
\(297x^{2} + 303y^{2} \frac{dy}{dx} = 0\)

\(303y^{2} \frac{dy}{dx} = -297x^{2},\)

\(\frac{dy}{dx} = -\frac{297x^{2}}{303y^{2}},\)

\(\frac{dy}{dx} = -\frac{99x^{2}}{101y^{2}}.\)

Given the equation \(101y^{3} = -99x^{3}\), we can simplify further by multiplying both sides by x:

\(101 \cdot 3y^{2} \frac{dy}{dx} = -99 \cdot 3x^{2},\)

\(101 \cdot y^{2} \frac{dy}{dx} = -99 \cdot x^{2},\)

Multiplying both sides by x:

\(101 \cdot x \cdot y^{2} \frac{dy}{dx} = -99 \cdot x^{3},\)

Given \(99x^{3} = -101y^{3}\):

\(\frac{dy}{dx} = \frac{101 y^{3}}{101 x y^{2}},\)

\(\frac{dy}{dx} = \frac{y}{x}.\)

So, the correct option is (D): \(\frac{y}{x}\)