Given y = $\mathit{a}{\mathit{e}}^{\mathit{b}\mathit{x}+5}$
Differentiating with y with respect to x,
⇒ $\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}$ = $\mathit{a}{\mathit{e}}^{\mathit{b}\mathit{x}+5}$ × b
⇒ $\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}$ = by ... (i) (Since y = $\mathit{a}{\mathit{e}}^{\mathit{b}\mathit{x}+5}$)
Diffrentiating (i) with respect to x,
⇒ $\frac{{\mathit{d}}^2\mathit{y}}{\mathit{d}{\mathit{x}}^2}$ = b $\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}$
⇒ $\frac{{\mathit{d}}^2\mathit{y}}{\mathit{d}{\mathit{x}}^2}$ = $\left(\frac{1}{\mathit{y}}\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}\right)\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}$ ... from (i)
⇒ y $\frac{{\mathit{d}}^2\mathit{y}}{\mathit{d}{\mathit{x}}^2}$ = ${\left(\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}\right)}^2$