Question

Hey, can anyone help me with this calculus problem

Differentiate the function.

y = e^(x + 2) + 4

y' = ?

Answer 1

\[\frac{ d(e ^{u}) }{ dx } = e ^{u} \times \frac{ du }{ dx }\]Just determine "u".

Answer 2

\[y = e ^{x} e ^{2} + 4\]
Now differentiate.

Answer 3

e^(x+2)

Answer 4

use the chain rule

Answer 5

Do you see that u = x + 2 ?

So, make the substitution and simplify.

Answer 6

Thanks for the replys everyone. I'm struggling with calculus and not totally sure how to do many problems. So boolan_abhi. This isn't about the power rule?

Answer 7

For functions where you have "e" to a given power, you just use the formula from my first post.

Answer 8

Power rule is used when it is of the form x power something.
Here we have e power something. So follow the first post of tcarrol010.

Answer 9

if u=g(x) is differentiable at x and y=f(u) is differentiable at u, then y is differentiable at x
i.e. dy/dx=(dy/du)*(du/dx)

Answer 10

It's actually using the chain rule, not the power rule, which is used for x power something, as @koushik_ksv said

Answer 11

One other thing, the second term is a constant , "4", so you can drop that right off if you want to because the derivative of a constant is "0".

Answer 12

okay. So here is what I have. I'm not sure if this is right though.

e^(x+2) * (x+2)/(dx) ? What ix dx

Answer 13

d(x+2)/(dx) not (x+2)/(dx)

d(x+2)/(dx) = 1

Answer 14

In words:

the derivative with respect to "x" of "x + 2" is 1.

Answer 15

I'm extremely confused with all these terms

Answer 16

When taking the derivative of "x + 2", you are taking the derivative of "x" plus the derivative of "2". The derivative of "2" is "0".

Answer 17

okay. i kinda get that. where does e come into this?

Answer 18

Have you heard of "e" before? It's the base for natural logs.

Answer 19

i heard of it before. but i don't understand the concept of it besides that it's a number

Answer 20

You are taking the derivative of an exponential function. It doesn't always have to be "e", but when it is the base, derivatives are greatly simplified for exponential functions.

Answer 21

derivative of e^x is e^x . WE can proof it usingfirst principle.
Let,
d(a^x)/dx=log a
put a=e
d(e^x)/dx=e^x*(ln e)=e^x

Answer 22

Okay, since derivative of e^x is e^x (using chain rule?) and constant is 0. Then the derivative would be e^x?

Answer 23

You are looking for the derivative of

e^(x + 2) not e^x

You don't have to re-create the wheel by expanding e^(x+ 2) into
(e^x)(e^2). That's wasted time. The formula has already been developed and it is much quicker just to use the formula from my first post which people who use calculus eventually commit to memory.

Answer 24

yup i go wid t carrol but its always wise to use first principles ..

Answer 25

In words, when you have "e" to some power, the derivative is that given "e" to some power TIMES the derivative of the exponent. That's all there is to it.

Answer 26

So, what's the exponent?

Answer 27

(x+2)?

Answer 28

Yes, and what is the derivative of that exponent. Just the exponent. In other words, what is the derivative of:

x + 2

?

Answer 29

If you don't know the derivative of:

x + 2

then I have to refer you to studying the basics of derivatives before trying to do the problems. You will have to know that first.