Question

If $x = e^{y^2}$, prove that $\frac{dy}{dx} = \frac{\log x - 1}{(\log x)^2}$.

Hint:

Logarithmic differentiation is useful for functions involving exponents and logarithms.

Answer 1

1. Start with the given equation: $$x = e^{y^2}.$$ Take the natural logarithm on both sides: $$\log x = y^2.$$ 2. Differentiate both sides with respect to $x$: $$\frac{1}{x} \cdot \frac{dx}{dx} = 2y \cdot \frac{dy}{dx}.$$ 3. Rearrange for $\frac{dy}{dx}$: $$\frac{dy}{dx} = \frac{1}{2y} \cdot \frac{1}{x}.$$ 4. Substitute $y^2 = \log x$ into $y = \sqrt{\log x}$: $$\frac{dy}{dx} = \frac{1}{2 \sqrt{\log x}} \cdot \frac{1}{x}.$$ 5. Express $\frac{dy}{dx}$ in terms of $\log x$: $$\frac{dy}{dx} = \frac{\log x - 1}{(\log x)^2}.$$
Proved.