Question

Find the second order derivative of the given functions.

$e^{6x}cos3x$

Answer 1

Let y = $e^{6x}cos3x$

Differentiating twicely w.r.t. x, we get

$\frac{dy}{dx}=\frac{d}{dx}\left(e^{6x}cos3x\right)=e^{6x}\frac{d}{dx}\left(cos3x\right)+cos3x\frac{d}{dx}(e{)}^{6x}=e^{6x}(-sin3x.3)+(cos3x)e^{6x}.6=e^{6}x\{-3sin3x+6cos3x\}and\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\{e^{6x}(-3sin3x+6cos3x)\}=e^{6x}\frac{d}{dx}(-3sin3x+6cos3x)+(-3sin3x+6cos3x)\frac{d}{dx}{e}^{6x}=e^{6x}(-3cos3x.3-6sin3x.3)+(-3sin3x+6cos3x)e^{6x}.6=e^{6x}\{-9cos3x-18sin3x-18sin3x+36cos3x\}=e^{6x}\{27cos3x-36sin3x\}=9e^{6x}\{3cos3x-4sin3x\}$