- #1

- 52

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**[SOLVED] first derivative: chain rule: easy for you guys**

Y=E^(-mx)

f= E^x g= -mx

f'= E^x g'= o

E^(-mx) * 0(E^(-mx))

i think, not sure though

Y'= 0

which is wrong

someone help

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- Thread starter physicsed
- Start date

- #1

- 52

- 0

Y=E^(-mx)

f= E^x g= -mx

f'= E^x g'= o

E^(-mx) * 0(E^(-mx))

i think, not sure though

Y'= 0

which is wrong

someone help

- #2

- 1,752

- 1

[tex]f(x)=e^{g(x)}[/tex]

[tex]f'(x)=e^{g(x)}g'(x)[/tex]

[tex]y=e^{-mx}[/tex]

[tex]y'=-me^{-mx}[/tex]

[tex]f'(x)=e^{g(x)}g'(x)[/tex]

[tex]y=e^{-mx}[/tex]

[tex]y'=-me^{-mx}[/tex]

Last edited:

- #3

- 52

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yes!!!!!!!!!!!!!!!!!

- #4

- 1,752

- 1

the derivative of e is itself ... times the derivative of it's exponent.

- #5

- 52

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how did you get g'(x)

i thought the (m) was constant thus making the the derivative equal to 0

i thought the (m) was constant thus making the the derivative equal to 0

- #6

- 1,752

- 1

Do this one for me ...

[tex]y=ax[/tex]

What is it's derivative?

[tex]y=ax[/tex]

What is it's derivative?

- #7

- 52

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well i would use the product rule or maybe not.

f= x g=ax

f'= 1 g'=product rule

high school calc.

- #8

- 1,752

- 1

The derivative of a constant times a function is simply ...

[tex]\frac{d}{dx}=cf'(x)[/tex]

[tex]y=ax[/tex]

[tex]y'=a[/tex] by constant rule

[tex]y'=a+x\cdot0=a[/tex] by product rule

Do you see why we don't need the product rule?

- #9

- 52

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[tex]ysingle-quote=a+x\cdot0=a[/tex]

times 0 would be zero,right?

times 0 would be zero,right?

- #10

- 52

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never mind, sorry.

- #11

- 52

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[tex]fsingle-quote(x)=e^{g(x)}gsingle-quote(x)[/tex]

what is g'(X) equall to

- #12

- 1,752

- 1

read my previous posts!!!

- #13

- 52

- 0

g'= -m

Solved.

thanks for the help

Solved.

thanks for the help

- #14

- 1,752

- 1

:) anytime

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