Question

If a function is given as $y=5{{e}^{7x}}+6{{e}^{-7x}}$ then show that $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=49y.$

Answer 1

We have been given that a function y is $5{{e}^{7x}}+6{{e}^{-7x}}.$ We are supposed to find the second derivative and then show that it is equal to 49y. So, first, we will find the first derivative. We know that $\dfrac{d\left( A+B \right)}{dx}=\dfrac{d\left( A \right)}{dx}+\dfrac{d\left( B \right)}{dx},$ and then we will use the result $\dfrac{d\left( {{e}^{ax}} \right)}{dx}={{e}^{ax}}.\dfrac{d\left( ax \right)}{dx}$ to simplify further. Once we get the first derivative, we will differentiate it again to find the second derivative. Then we will take ${{7}^{2}}$ out and find our required solution.

Complete step-by-step answer:
We have y given as $y=5{{e}^{7x}}+6{{e}^{-7x}}$ and we have to double differentiate y to verify that $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=49y.$
Let us consider $y=5{{e}^{7x}}+6{{e}^{-7x}}$
Now, taking derivative, we have
$\dfrac{d\left( y \right)}{dx}=\dfrac{d\left( 5{{e}^{7x}}+6{{e}^{-7x}} \right)}{dx}$
So, applying the rule $\dfrac{d\left( A+B \right)}{dx}=\dfrac{d\left( A \right)}{dx}+\dfrac{d\left( B \right)}{dx},$ to the above, we have
$\Rightarrow \dfrac{d\left( y \right)}{dx}=\dfrac{d\left( 5{{e}^{7x}} \right)}{dx}+\dfrac{d\left( 6{{e}^{-7x}} \right)}{dx}$
Taking constants outside from each term, we get,
$\Rightarrow \dfrac{dy}{dx}=5\dfrac{d\left( {{e}^{7x}} \right)}{dx}+6\dfrac{d\left( {{e}^{-7x}} \right)}{dx}$
We know that, $\dfrac{d\left( {{e}^{z}} \right)}{dx}={{e}^{z}}\dfrac{dz}{dx}.$
So, we can write the derivative as $\dfrac{d\left( {{e}^{7x}} \right)}{dx}={{e}^{7x}}\dfrac{d\left( 7x \right)}{dx}$
$\Rightarrow \dfrac{d\left( {{e}^{7x}} \right)}{dx}={{e}^{7x}}.7.....\left( i \right)$
Similarly, we have
$\dfrac{d\left( {{e}^{-7x}} \right)}{dx}={{e}^{-7x}}\dfrac{d\left( -7x \right)}{dx}$
$\Rightarrow \dfrac{d\left( {{e}^{7x}} \right)}{dx}={{e}^{-7x}}\left( -7 \right).....\left( ii \right)$
Putting the above values from (i) and (ii) in derivative, we get,
$\Rightarrow \dfrac{dy}{dx}=5\times 7\times {{e}^{7x}}+6\times \left( -7 \right)\times {{e}^{-7x}}$
Taking 7 as common, we get
$\Rightarrow \dfrac{dy}{dx}=7\left( 5{{e}^{7x}}-6{{e}^{-7x}} \right)$
We have got the first derivative. Now we have to differente it again to get the second derivative. So, we get,
$\Rightarrow \dfrac{d\left( \dfrac{dy}{dx} \right)}{dx}=\dfrac{d\left( 7\left( 5{{e}^{7x}}-6{{e}^{-7x}} \right) \right)}{dx}$
Taking the constant, i.e 7 outside, we get
$\Rightarrow \dfrac{{{d}^{2y}}}{d{{x}^{2}}}=7\dfrac{d\left( 5{{e}^{7x}}-6{{e}^{-7x}} \right)}{dx}$
Simplifying using rule, we get,
$\Rightarrow \dfrac{{{d}^{2y}}}{d{{x}^{2}}}=7\left[ 5\dfrac{d\left( {{e}^{7x}} \right)}{dx}-6\dfrac{d\left( {{e}^{-7x}} \right)}{dx} \right]$
Using (i) and (ii), we get,
$\Rightarrow \dfrac{{{d}^{2y}}}{d{{x}^{2}}}=7\left[ 5\left( 7 \right){{e}^{7x}}-6\left( -7 \right){{e}^{-7x}} \right]$
Again, taking 7 common, we have
$\Rightarrow \dfrac{{{d}^{2y}}}{d{{x}^{2}}}=7\left[ 5\left( 7 \right){{e}^{7x}}+6\left( 7 \right){{e}^{-7x}} \right]$
Taking 7 out from both the terms, we get,
$\Rightarrow \dfrac{{{d}^{2y}}}{d{{x}^{2}}}={{7}^{2}}\left[ 5{{e}^{7x}}+6{{e}^{-7x}} \right]$
We know that, ${{7}^{2}}=49$ and $5{{e}^{7x}}+6{{e}^{-7x}}=y.$ So, we get,
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=49y$
Hence we have proved the result.

Note: We should note that $\dfrac{d\left( {{e}^{-7x}} \right)}{dx}\ne {{e}^{-7x}}.$ Here we have to apply the product rule.
$\dfrac{d\left( {{e}^{-7x}} \right)}{dx}=\left( {{e}^{-7x}} \right)\times \dfrac{d\left( -7x \right)}{dx}$
$\Rightarrow \dfrac{d\left( {{e}^{-7x}} \right)}{dx}={{e}^{-7x}}\times \left( -7 \right)$
$\Rightarrow \dfrac{d\left( {{e}^{-7x}} \right)}{dx}=-7{{e}^{-7x}}$
Whenever the power or the exponential has power different than x, then we have to take the derivative separately for a clear solution.