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Q)

If $ y = 500e^{\large 7x} + 600e^{\large -7x}$, what is $\large\frac{d^2y}{dx^2}$

$\begin{array}{1 1} 49y \\ 61y \\ 51y \\ 39 y \end{array} $

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A)
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^{\large x})=e^x$
Step 1:
We have $y=500e^{\large 7x}+600e^{\large-7x}$
Differentiating with respect to $x$
$\large\frac{dy}{dx}$$=500.7e^{\large 7x}+600.(-7)e^{\large -7x}$
Step 2:
$\large\frac{d^2y}{dx^2}$$=500\times 7\times 7e^{\large 7x}+600(-7)\times (-7)e^{\large-7x}$
$\quad\;=500e^{\large 7x}.49+600e^{\large -7x}.49$
$\quad\;=49[500e^{\large 7x}+600e^{-7x}]$
$\quad\;=49y$
Hence proved.
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