Logarithmic Differentiation

Some functions are awkward to differentiate directly — long products, quotients or powers of several factors — or are of the genuinely new form $y=[f(x)]^{g(x)}$, where both the base and the exponent contain $x$; here neither the ordinary power rule (which needs a constant exponent) nor the ordinary exponential rule (which needs a constant base) applies on its own. Logarithmic differentiation solves both problems by taking the natural logarithm of both sides first: $\log y=g(x)\log[f(x)]$, which the laws of logarithms turn into a sum instead of a product, and a multiple instead of a power — far easier to differentiate term by term. Differentiating implicitly with respect to $x$ gives $\dfrac1y\dfrac{dy}{dx}=$ (derivative of the right side), so $\dfrac{dy}{dx}=y\times$(that derivative), with $y$ finally substituted back as the original expression $[f(x)]^{g(x)}$.

Worked Example 1 — differentiate w.r.t. x:

1. $y=\dfrac{(x^2+3)^2(x^3+5)^{2/3}}{(2x^2+1)^{3/2}}$: taking logs, $\log y=2\log(x^2+3)+\frac23\log(x^3+5)-\frac32\log(2x^2+1)$. Differentiating implicitly and multiplying back by $y$: $\dfrac{dy}{dx}=y\left[\dfrac{4x}{x^2+3}+\dfrac{2x^2}{x^3+5}-\dfrac{6x}{2x^2+1}\right].$
2. $y=\dfrac{e^{x^2}(\tan x)^{x/2}}{(1+x^2)^{3/2}(\cos x)^3}$: taking logs, $\log y=x^2+\frac x2\log(\tan x)-\frac32\log(1+x^2)-3\log(\cos x)$. Differentiating and multiplying back: $\dfrac{dy}{dx}=y\left[2x+\dfrac{x}{2\sin x\cos x}+\dfrac12\log(\tan x)-\dfrac{3x}{1+x^2}+3\tan x\right].$
3. $y=(x+1)^{3/2}(2x+3)^{5/2}(3x+4)^{2/3}$: taking logs and differentiating similarly, $\dfrac{dy}{dx}=y\left[\dfrac3{2(x+1)}+\dfrac{15}{2(2x+3)}+\dfrac2{3x+4}\right].$
4. $y=x^a+x^x+a^x$: split into $u=x^a+a^x$ (ordinary power/exponential rules apply directly) and $v=x^x$ (needs log-differentiation, since both base and exponent are $x$): $\dfrac{du}{dx}=ax^{a-1}+a^x\log a$; for $v=x^x$, $\log v=x\log x$, so $\dfrac1v\dfrac{dv}{dx}=\log x+1$, giving $\dfrac{dv}{dx}=x^x(1+\log x)$. Adding: $\dfrac{dy}{dx}=ax^{a-1}+a^x\log a+x^x(1+\log x).$ …