Higher-Order Derivatives and Choosing the Right Rule

Higher-order derivatives

Differentiating $f'$ again gives the second derivative $f''$; repeating gives $f'''$, $f^{(4)}$, … In Leibniz notation $\dfrac{d^2y}{dx^2}$, $\dfrac{d^ny}{dx^n}$. Nothing new is memorised — but every product, quotient or chain structure returns at every step, so simplify and factor before differentiating again.

Worked example: $y=x^2e^{3x}$. Product and chain: $y'=2xe^{3x}+3x^2e^{3x}=e^{3x}(3x^2+2x)$. Product again: $y''=3e^{3x}(3x^2+2x)+e^{3x}(6x+2)=e^{3x}(9x^2+12x+2)$. Factoring $e^{3x}$ out first is what keeps the second step short.

Implicit curves. $\dfrac{d^2y}{dx^2}$ is the derivative of the expression for $\dfrac{dy}{dx}$ (quotient rule, $y$ still a function of $x$), after which the known $\dfrac{dy}{dx}$ must be substituted back so the answer contains only $x$ and $y$.

Choosing the procedure

Rewrite first: $\sqrt{x}=x^{1/2}$, $\dfrac{1}{x^3}=x^{-3}$, $\dfrac{x^2+1}{x}=x+x^{-1}$ — many quotient-rule problems are power-rule problems in disguise. Bases other than $e$: $a^x=e^{x\ln a}$, hence $\dfrac{d}{dx}a^x=a^x\ln a$ and $\dfrac{d}{dx}\log_a x=\dfrac{1}{x\ln a}$.

Cyclic patterns. Derivatives of $\sin x$ cycle $\cos x,\ -\sin x,\ -\cos x,\ \sin x$ with period $4$, so for $f(x)=\sin x$, $f^{(27)}(x)=f^{(3)}(x)=-\cos x$ because $27=4\cdot6+3$.