Question

Where is the second derivative of y = 2xe−x equal to 0?

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Answer 1

@ParthKohli

Answer 2

\[\large\rm y=2xe^{-x}\]Do you understand how to find the first derivative? :)

Answer 3

Use the product rule:

d/dx (f•g) = f'•g + f•g'

Answer 4

Do you want an example (with some notes and explanations)?

Answer 5

I got -2e^-x(x-1)

Answer 6

For first derivative? Looks good so far :)

Answer 7

the first, that is right...

Answer 8

okay good. now what?

Answer 9

second derivative

Answer 10

differentiate the result (that you obtained for the first derivative).

This wll give you the 2nd derivative.

Answer 11

(-2e^-x(x-3) ??

Answer 12

oh no didn't you say it would be 2e^-x(x-2)

Answer 13

yes, checked it, correct

Answer 14

but you should differentiate it yourself, instead of saying; wouldn't you want to do it yourself (tho)?

Answer 15

\(\large\color{#000000 }{ \displaystyle y^{(0)}(x)=kxe^{bx} }\)

\(\large\color{#000000 }{ \displaystyle y^{(1)}(x)=ke^{bx}+bkxe^{bx} }\)

\(\large\color{#000000 }{ \displaystyle y^{(2)}(x)=ke^{bx}+bke^{bx}+b^2kxe^{bx} }\)
\(\large\color{#000000 }{ \displaystyle y^{(2)}(x)=(k+bk)e^{bx}+b^2kxe^{bx} }\)

Answer 16

in general, (with chain rule to the exponent, and the product rule of course)

Answer 17

In any case, you have to set f''(x)=0 and solve for x.

Answer 18

2e^-x(x-2)=0

Answer 19

2e^(-x)\(\ne\)0 \(\forall\)x

Answer 20

So, (x-2)=0

good luck!