Question

Find the second order derivatives of the function.e^6x.cos3x

Answer 1

Let $\bf{y=e^{6x}cos3x}$
now difference y with respect to x ,
$\bf{\frac{dy}{dx}=\frac{d(e^{6x}cos3x)}{dx}}\\\\=\bf{e^{6x}\frac{d(cos3x)}{dx}+cos3x\frac{d(e^{6x})}{dx}}\\\\=\bf{e^{6x}.(-3sin3x)+cos3x.6e^{6x}}\\\\=\bf{e^{6x}(-3sin3x+6cos3x)}$

so, $\frac{dy}{dx}=e^{6x}(-3sin3x+6cos3x)$
now differentiate $\frac{dy}{dx}$ with respect to x once again,

$\bf{\frac{d^2y}{dx^2}=\frac{d\{e^{6x}(-3sin3x+6cos3x)\}}{dx}}\\\\=\bf{e^{6x}\{(-9cos3x)+(-18sin3x)\}+6e^{6x}(-3sin3x+6cos3x)}\\\\=\bf{e^{6x}(27cos3x-36sin3x)}$

hence, d²y/dx² = 9e^{6x}(3cos3x - 4sin3x)