Q.Differentiate the following w.r.t. $x$: $\dfrac{4x-1}{(2x+3)(5-2x)^2}$

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

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

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

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 $y'=\tfrac{dy}{dx}$:

$\frac{m}{x} + \frac{n}{y}y' = \frac{(m+n)}{x+y} + \frac{(m+n)}{x+y}y'.$

The algebra simplifies (the $m$ and $n$ terms recombine) to the well-known result

$\frac{dy}{dx} = \frac{y}{x}.$

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):

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

Step 3 — solve for $dy/dx$:

$\frac{dy}{dx} = y(2\ln x + 2) = x^{2x} \cdot 2(\ln x + 1) = 2x^{2x}(1 + \log x)$