Question

What is the second derivative of #f(x)=e^x + e^(-x / 2 #?

Answer 1

`d^2/dx^2 f(x) = f''(x) = d/dx(f'(x))`

the first derivative by using the chain rule is
`d/dx f(x) = e^x - e^(-x/2)/2`

then take the derivative of that function

therefore,
again using the chain rule on the second element,

`d/dx f'(x) = e^x + e^(-x/2)/4`
therefore the answer is

`f'(x)=`
`e^x + e^(-x/2)/4`
=

Answer 2

`color(blue)(e^x+1/4e^(-1/2x))`

Explanation:

Since the second derivative is the derivative of the first derivative, we start by finding:

`dy/dx(e^x+e^(-x/2))`

Re-writing:

`e^(-x/2)=(e^x)^(-1/2)`

We know `dy/dxe^x=e^x`

We need to use the Chain Rule for `(e^x)^(-1/2)`

`dy/dx(e^x+(e^x)^(-1/2))=e^x-1/2(e^x)^(-3/2)*e^x`

`=e^x-1/2e^(-3/2x+x)=e^x-1/2e^(-1/2x)`

We now differentiate this again:

Re-writing as before:

`e^x-1/2e^(-1/2x)=e^x-1/2(e^x)^(-1/2)`

`dy/dx(e^x-1/2(e^x)^(-1/2))=e^x+1/4(e^x)^(-3/2)*e^x`

`=e^x+1/4e^(-3/2x+x)=e^x+1/4e^(-1/2x)`

`:.`

`(d^2y)/(dx^2)(e^x+e^(-x/2))=color(blue)(e^x+1/4e^(-1/2x))`