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

2 Answers
Mar 12, 2018

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
=

Mar 12, 2018

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))