Q.If $y = e^{2x}$, find $\dfrac{d^2 y}{dx^2}$ and hence show that $\dfrac{d^2 y}{dx^2} = 4y$.

Differentiate twice using the chain rule, then compare the result with $4y$.

Differentiate $y = e^{2x}$ twice, using $\dfrac{d}{dx}e^{u} = e^{u}\dfrac{du}{dx}$ with $u = 2x$ (chain rule, §3).

First derivative. $\dfrac{dy}{dx} = e^{2x}\cdot 2 = 2e^{2x}$.

Second derivative. Differentiate again: $\dfrac{d^2 y}{dx^2} = \dfrac{d}{dx}\big(2e^{2x}\big) = 2\cdot e^{2x}\cdot 2 = 4e^{2x}$.

Show the required relation. Since $y = e^{2x}$, we have $4e^{2x} = 4y$. Therefore

as required. (Equivalently, $\dfrac{d^2 y}{dx^2} - 4y = 0$.) …

The main errors are forgetting the chain-rule factor $2$ at each stage (writing $\tfrac{dy}{dx}=e^{2x}$ instead of $2e^{2x}$, which would wrongly give $\tfrac{d^2y}{dx^2}=e^{2x}=y$), and confusing $\tfrac{d^2y}{dx^2}$ with $ …