Question

find the derivative of y=xe^(x)

Answer 1

use the product rule

Answer 2

\(\large\color{black}{ \frac{d}{dx}~\left[~f(x){\color{white}{\normalsize\int}}g(x)~\right]~=~f'(x)g(x)+f(x)g'(x)}\)

Answer 3

that makes sense now, thank you!

Answer 4

(Or, if you don't want to go by the product rule, you can use logarithmic differentiation, but I don't think you want that)

Answer 5

no i think product rule would be the easiest thanks :)

Answer 6

yes, it would be

Answer 7

(logarithmic rule is just 1 way to prove product and quotient rules)

Answer 8

(I mean log. differentiation, not log. rule)

Answer 9

what is the derivative of x?
what is the derivative of e^x?

Answer 10

You can use the product rule, the answer is y'=e^(x)+xe^x

Answer 11

derivative of x is 1 and derivative of e^x is e^x so then the answer would be xe^x+e^x

Answer 12

d/dx of e^x = e^x

Answer 13

yes

Answer 14

\(\large\color{black}{ \displaystyle y=xe^x }\)
\(\large\color{black}{ \displaystyle y'=e^x+xe^x }\)

Answer 15

and then if you differentiate again, you will see
\(\large\color{black}{ \displaystyle y=xe^x }\)
\(\large\color{black}{ \displaystyle y'=e^x+xe^x }\)
\(\large\color{black}{ \displaystyle y''=e^x+e^x+xe^x=2e^x+xe^x }\)
\(\large\color{black}{ \displaystyle y'''=2e^x+e^x+xe^x=3e^x+xe^x }\)
\(\large\color{black}{ \displaystyle y''''=3e^x+e^x+xe^x=4e^x+xe^x }\)

and on....

Answer 16

you can observe, that

\(\Large\color{black}{ \displaystyle y=xe^x }\)

then, for any Ath derivative
\(\Large\color{black}{ \displaystyle \frac{d^{\color{red}{\rm {\tiny~}a}}y}{dx^{\color{red}{\rm {\tiny~}a}}}={\color{red}{\rm a{\tiny~}}}e^x+xe^x }\)

Answer 17

are you cool with that?

Answer 18

yeah i get it thanks!

Answer 19

Well, for notation sake,

\(\Large\color{black}{ \displaystyle {\color{red}{\rm a{\tiny~}}}\in{\bf N} }\)

Answer 20

saying,
a is a natural number.

Answer 21

yw....