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Find the second order derivatives of \( e^{6x} \cos 3x\)

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Toolbox:
  • $y=f(x)$
  • $\large\frac{dy}{dx}$$=f'(x)$
  • $\large\frac{d^2y}{dx^2}=\frac{d}{dx}\big(\frac{dy}{dx}\big)$
  • $\large\frac{d}{dx}$$e^x=e^x$
Step 1:
Let $e^{\large 6x}\cos 6x$
Differentiating with respect to $x$
$\large\frac{dy}{dx}$$=e^{6x}\large\frac{d}{dx}$$(\cos 3x)+\cos 3x\large\frac{d}{dx}$$e^{6x}$
$\quad\;=e^{6x}(-3\sin 3x)+\cos 3x.6.e^{6x}$
$\quad\;=e^{6x}(-\sin 3x+6\cos 3x)$
Step 2:
$\large\frac{d^2y}{dx^2}$$=e^{6x}\large\frac{d}{dx}$$(-3\sin 3x+6\cos 3x)+(-3\sin 3x+6\cos 3x)\large\frac{d}{dx}$$(e^{6x})$
$\quad\;=e^{6x}(-9\cos 3x-18\sin 3x)+(-3\sin 3x+6\cos 3x.e^{6x}.6)$
$\quad\;=9e^{6x}(3\cos 3x-4\sin 3x)$
answered May 13, 2013 by sreemathi.v
 
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