Question

Need help proving that the second derivative of y=a/2(e^(x/a)+e^(-x/a)) is y/a^2 through differentiation. Thanks

Answer 1

Well if you take the derivative of an exponential function, what happens? Taking derivative of e^(kx), you just get e^(kx) * k, right?

So what we could do is differentiate the current function twice using exponents rule. Then factor some constants out to get the statement for y, along with the constant multiple 1/a^2 that we factor out.

Answer 2

can you please help me with the first differentiation and i will then do the second

Answer 3

Sure. To make it easier we might distribute a/2 to both terms first...
\( y = \dfrac{a}{2} \left( e^{x/a} + e^{-x/a} \right) = \dfrac{a}{2} e^{x/a} + \dfrac{a}{2} e^{-x/a} \)

Answer 4

Now when we take the derivative of both sides to find first derivative, we simply take the derivative term-by-term:
Derivative of \( \dfrac{a}{2} e^{x/a} \). Note that the exponent \(x/a\) is just x multiplied by a constant, \(1/a\). So we use derivative of exponent \(\dfrac{d}{dx} e^{kx} = k e^{kx} \)
\( \dfrac{d}{dx} \left( \dfrac{a}{2} e^{x/a} \right) = \dfrac{a}{2} e^{x/a} \times \dfrac{1}{a} \)

Is any of this unclear so far?

Answer 5

no i understand this thank you

Answer 6

Alright. Because every step after this is virtually identical.
The derivative of the second term is nearly the same as the first, only our constant is -1/a instead of 1/a.
\( \dfrac{d}{dx} \left( \dfrac{a}{2} e^{-x/a} \right) = \dfrac{a}{2} e^{-x/a} \times \left(-\dfrac{1}{a}\right) \)

Answer 7

yes as it is just the difference of the negative

Answer 8

Giving us the first derivative:
\(y' = \dfrac{a}{2} e^{x/a} \times \dfrac{1}{a} + \dfrac{a}{2} e^{-x/a} \times \dfrac{-1}{a} \)
You could simplify but what we end up doing is factoring out the 1/a^2 again to show that it is y/a^2. :)

Answer 9

Factoring out the 1/a^2 after we take the second derivative, that is *

Answer 10

ok

Answer 11

Were you able to figure it out from there?

Answer 12

i beleive so

Answer 13

Alright. Essentially if you obtained the result:
\(y''= \dfrac{a}{2} e^{x/a} \times \dfrac{1}{a} \dfrac{1}{a} + \dfrac{a}{2} e^{-x/a} \times \dfrac{-1}{a} \dfrac{-1}{a} \)
\(y''= \dfrac{1}{a^2} \underbrace{\left( \color{blue}{\dfrac{a}{2} e^{x/a} + \dfrac{a}{2} e^{-x/a}}\right) }_{y} \)
\( y''= y / a^2 \)
You would be golden. :)

Answer 14

thank you so much i almost had it

Answer 15

Glad to help! :)