Question

prime question:
f(x)=2e^(x+2)+e^a
f`(x)=

Answer 1

mistake: e^a=> e^9

Answer 2

f`(x)=2e^(x+2)

Answer 3

\[f(x)=2e^{x+2}+e^9\]\[f'(x)=2e^{x+2}\]

Answer 4

how did you get this answer?

Answer 5

\[f(x)=2e^{x+2}+e^9\]
\[d(2e^{x+2})/dx=2\ d(e^{x+2})/dx\]

Answer 6

u = x+2, du = 1 dx\[e^{x+2}=e^u=du/dx\ e^u=1e^{x+2}\]

Answer 7

I entered the answer and it told me that it was wrong

Answer 8

No i am sorry! f`(0)=

Answer 9

no x!

Answer 10

So plug in 0

Answer 11

You need to get your stuff together, dodo1

Answer 12

into x?

Answer 13

f'(x) is a function. So yes, f'(0) would be f'(x) with 0 for x.

Answer 14

so is 2?

Answer 15

no its 2e^2?

Answer 16

\[f'(x)=2e^{x+2}\]\[f'(0)=2e^{(0)+2}\]

Answer 17

Thank you so much!

Answer 18

how do you sqrt prime?
let say f(x)=sqrt(4x)

Answer 19

the final answer is 4x1/2?

Answer 20

u = 4x

du/dx = 4

\[d(\sqrt {u})/du = \frac{1}{2\sqrt u}du/dx\]\[\frac{1}{2\sqrt {4x}}4=\frac{2}{\sqrt{4x}}=\frac{2}{2\sqrt x}=\frac{1}{\sqrt x}\]

Answer 21

Ok,the final answer is 1/sqrt(x)