Q.Find $\dfrac{dy}{dx}$ if $y = \dfrac{x^2 (2x+1)^3}{\sqrt{x+4}}$.

This is a long product/quotient, so take logs to turn it into a sum before differentiating.

The expression is a product and quotient of several factors; logarithmic differentiation (§5) turns it into a manageable sum.

Take natural logs and expand using $\ln(ab)=\ln a+\ln b$, $\ln\tfrac{a}{b}=\ln a-\ln b$, $\ln(a^n)=n\ln a$ (and $\sqrt{x+4}=(x+4)^{1/2}$):

Differentiate both sides with respect to $x$. The left side gives $\dfrac1y\dfrac{dy}{dx}$. On the right, each $\ln u$ term differentiates to $\dfrac{u'}{u}$:

Solve for $\dfrac{dy}{dx}$ by multiplying by $y$ and substituting the original expression:

$\frac{dy}{dx} = \frac{x^2 (2x+1)^3}{\sqrt{x+4}}\left[\frac{2}{x} + \frac{6}{2x+1} - \frac{1}{2(x+4)}\right].$ …

The same derivative could be found by combining the product and quotient rules directly, but with three factors that is error-prone; logs reduce it to differentiating a sum of three simple terms — the reason logarithmic differen …

Common errors: dropping the coefficient when using $\ln(a^n)=n\ln a$ (e.g. writing $\ln(2x+1)$ instead of $3\ln(2x+1)$); mishandling the square root's exponent ($\tfrac12$, and negative because it is in the denominator, giving $-\tfr …