Question

If \(x^{\frac{3}{4}}+y^{\frac{3}{4}}=a^{\frac{3}{4}}\) (a is a constant) then \(\frac{dy}{dx}\) is equal to :

• A. \((\frac{y}{x})^{\frac{1}{4}}\)
• B. \((\frac{x}{y})^{\frac{1}{4}}\)
• C. \(-(\frac{y}{x})^{\frac{1}{4}}\)
• D. \(-(\frac{x}{y})^{\frac{1}{4}}\)

Answer 1

The Correct Option is C

To solve for \(\frac{dy}{dx}\) given the equation \(x^{\frac{3}{4}}+y^{\frac{3}{4}}=a^{\frac{3}{4}}\), we must differentiate both sides with respect to \(x\).

Start with the original equation:

\(x^{\frac{3}{4}}+y^{\frac{3}{4}}=a^{\frac{3}{4}}\)

Differentiating each term with respect to \(x\), we have:

\(\frac{d}{dx}(x^{\frac{3}{4}}) + \frac{d}{dx}(y^{\frac{3}{4}}) = \frac{d}{dx}(a^{\frac{3}{4}})\)

Since \(\frac{d}{dx}(a^{\frac{3}{4}})=0\) (as \(a\) is a constant), this simplifies to:

\(\frac{3}{4}x^{-\frac{1}{4}} + \frac{3}{4}y^{-\frac{1}{4}}\frac{dy}{dx} = 0\)

Rearranging to solve for \(\frac{dy}{dx}\), we find:

\(\frac{3}{4}y^{-\frac{1}{4}}\frac{dy}{dx} = -\frac{3}{4}x^{-\frac{1}{4}}\)

Dividing both sides by \(\frac{3}{4}y^{-\frac{1}{4}}\) gives:

\(\frac{dy}{dx} = -\frac{x^{-\frac{1}{4}}}{y^{-\frac{1}{4}}}\)

Simplifying the expression, we have:

\(\frac{dy}{dx} = -(x^{-\frac{1}{4}}y^{\frac{1}{4}}) = -(\frac{y}{x})^{\frac{1}{4}}\)

Thus, the correct answer is \(-(\frac{y}{x})^{\frac{1}{4}}\).