Question

f(x)=2x^2e^5x+1 find derivative

Answer 1

@Mcjay Are you familiar with the chain rule? And welcome to OpenStudy :-D

Answer 2

Chain Rule:
for:
\[u^n \]
then:
\[(n*u^{n-1})*\frac{d}{du}u\]
Do the power rule, then take the derivative of what's inside the "shell"/"layer"/"whatever"

Answer 3

basically i'm looking how i can getlocal max and min

Answer 4

after derivative i find this i dont know if is correct 20 x e^5x

Answer 5

Local max & min will occur at critical points.

To find critical points you need to find the first derivative, then find what the input needs to be to make the output (the slope) be zero (be horizontal).

f(x) = y
f'(x) = y' = \(\large\frac{dy}{dy}\)

When y' = 0 , you'll have the critical points.


Write up what you think the derivative is into the equation editor (click the "\(\Sigma\) Equation" button below) , you can write fractions by doing this:

frac{dy}{dx} turns into \(\large\frac{dy}{dy}\)

Answer 6

\[f(x) = 2 x^2 e^5 x+1\]
\[f'(x)=\frac{d}{dx}(2 x^2 e^5 x+1)\]
\[f'(x)=2 e^5 (\frac{d}{dx}(x^3))+\frac{d}{dx}(1)\]
\[f'(x)=2 e^5 (3x^2)+0 \ \ \ \ \ dx\]
\[f'(x)=6 e^5 x^2 \ \ \ \ \ \ dx\]

Answer 7

\[dy/dx 2x^2e ^(5x) +1

Answer 8

5x

Answer 9

Actually I just noticed I had a typo, the derivative of y is dy, so y' = dy, not dy/dy. Derp.

Answer 10

right

Answer 11

Minor detail, but just to point that out if you're writing this down for your notes.

Answer 12

i try to write e ^(5x) but i dont get it

Answer 13

you can divide both sides by dx, to just get dy/dx which of course IS your instantaneous slope.



Average slope:
\[m=\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\]

Instantaneous slope:
\[m=\frac{dy}{dx}\]

Answer 14

Oh you mean what you intended to write wasn't \(2x^2e^5x+1\)?

Answer 15

was not e is exponent 5x

Answer 16

Yeah you REALLY got to watch your parenthesis here, especially when asking question online. We can't see your original problem of course, so you got to type it in to make sure you communicate clearly.

\[\frac{d}{dx}(2 x^2 e^{5 x}+1) \ ????\]

Answer 17

yes

Answer 18

sorry about that it is my first time i use online help

Answer 19

help please

Answer 20

Factor out the two, apply the derivative to both terms (it's associative):
\[2 (\frac{d}{dx}(e^{5 x} x^2))+\frac{d}{dx}(1)\]
Using the product rule, "first d second + second d first":
\[2(x^2 (\frac{d}{dx}(e^{5 x}))+e^{5 x} (\frac{d}{dx}(x^2)))\]
Using the chain rule:
\[2(x^2 (e^{5 x} (\frac{d}{dx}(5 x)))+e^{5 x} (\frac{d}{dx}(x^2)))\]

Answer 21

Can you finish from here? You'll have this mess simplify to something like this (if you factor out the common term):
\[2xe^{5 x}(5 x+2) = \frac{dy}{dx} = m = slope\]

You need to find what makes the slope zero. What the "x" input needs to be.

Answer 22

5x+2 =0

Answer 23

Yep, that's one of the roots, now what's the other? :-)

Answer 24

x=-2/5

Answer 25

Anything times zero is...?

Answer 26

is 0

Answer 27

i think to find local min and max u find x first

Answer 28

Yep, that's the other.

Kind of hard to see in the original graph but that's the two times it reaches zero.