Question

how do you find the derivative of:
f(x)=4e^(x+3)+e^1

Answer 1

e^1 +4*e^(x+3)

Answer 2

f(x)=4e^(x+3)+e^1

Well, you're going to need to know five things.
1: d/dx [e^x] = e^x
2: d/dx [f(g(x))] = f'(g(x)) * g'(x), otherwise known as the chain rule.
3: d/dx [f(x) + g(x)] = d/dx [f(x)] + d/dx [g(x)]
4: d/dx [c * f(x)] = c * d/dx [f(x)], where c is any constant.
5: d/dx [c] = 0, the derivative of any constant is 0.

Knowing those four things, finding the derivative is a snap.


First, use 4 and 3 to get this.
f'(x) = 4 * d/dx [e^(x+3)] + d/dx [e^1]
Now, 5, 1 and 2.
f'(x) = 4 * e^(x+3) + 0
Simplify.

f'(x) = 4e^(x+3)

Answer 3

akabhinav32 is incorrect. e^1 = e, and the derivative of any constant is 0.

Answer 4

thank you!

Answer 5

oh and also if y=e^(g(x)) then y'=g'(x)*e^(g(x))

Answer 6

Yeah. That's just the chain rule, really.