Q.Find $\frac{dy}{dx}$, if $y = x^x$, $x > 0$.

$\frac{dy}{dx} = x^x(\log x + 1)$.

Take logs of $y = x^x$:

$\log y = x\log x.$

Differentiate implicitly:

$\frac{1}{y}\frac{dy}{dx} = \log x + x\cdot\frac{1}{x} = \log x + 1.$

So …

Use logarithmic differentiation on $x^m y^n=(x+y)^{m+n}$; the cross terms cancel and $\dfrac{dy}{dx}=\dfrac{y}{x}$.

Logarithmic differentiation of an implicit relation is a standard NCERT/CBSE Class 12 continuity and differentiability method.

Take natural logs of both sides:

$m\ln x + n\ln y = (m+n)\ln(x+y).$

Differentiate with respect to $x$:

$\frac{m}{x} + \frac{n}{y}\frac{dy}{dx} = (m+n)\cdot\frac{1+\tfrac{dy}{dx}}{x+y}.$

Multiply out and collect. Writing $y'=\tfrac{dy}{dx}$: …

When the variable appears in both the base and the exponent, take logarithms first (logarithmic differentiation), then differentiate implicitly.

Given: $y = x^{2x}$

Step 1 — take log of both sides:

$\ln y = 2x \ln x$

Step 2 — differentiate both sides w.r.t. $x$ (product rule on the right):

$\frac{1}{y}\frac{dy}{dx} = 2\ln x + 2x \cdot \frac{1}{x} = 2\ln x + 2$

…