Question

If $y^3=x$ then the value of $\frac{dy}{dx}$ at $x=1$ is

• A. -3
• B. 3
• C. $\frac{1}{3}$
• D. $\frac{1}{\sqrt{3}}$

Hint:

For implicit differentiation, always remember to multiply by $\frac{dy}{dx}$ whenever you differentiate a term containing $y$. After differentiating, you'll have an equation involving $x, y,$ and $\frac{dy}{dx}$. Solve for $\frac{dy}{dx}$ algebraically. If you need to evaluate the derivative at a specific point, you'll often need to find the coordinates of both $x$ and $y$ at that point.

Answer 1

The Correct Option is C

Step 1: Find y at x=1: Substitute \( x=1 \) into \( 3^x y^x = x^{3y} \): \( 3^1 \cdot y^1 = 1^{3y} \implies 3y = 1 \implies y = \frac{1}{3} \).
Step 2: Logarithmic Differentiation: Take log of both sides: \( \ln(3^x y^x) = \ln(x^{3y}) \). \( x\ln 3 + x\ln y = 3y \ln x \). Differentiate w.r.t \( x \): \[ \ln 3 + \left(1 \cdot \ln y + x \cdot \frac{y'}{y}\right) = 3\left(y' \ln x + y \cdot \frac{1}{x}\right) \]
Step 3: Substitute values: Put \( x=1, y=1/3 \). Note \( \ln 1 = 0 \). \[ \ln 3 + \ln(1/3) + 1 \cdot \frac{y'}{1/3} = 3\left(0 + \frac{1/3}{1}\right) \] \[ \ln 3 - \ln 3 + 3y' = 3(1/3) \] \[ 3y' = 1 \implies y' = \frac{1}{3} \]