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First derivative: chain rule: easy for you guys

  • Thread starter physicsed
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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
 

Answers and Replies

1,750
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]
 
Last edited:
52
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yes!!!!!!!!!!!!!!!!!
 
1,750
1
the derivative of e is itself ... times the derivative of it's exponent.
 
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
 
1,750
1
Do this one for me ...

[tex]y=ax[/tex]

What is it's derivative?
 
52
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gosh!
well i would use the product rule or maybe not.
f= x g=ax
f'= 1 g'=product rule
high school calc.
 
1,750
1
Don't worry about the product rule when you're differentiating with a constant.

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?
 
52
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[tex]ysingle-quote=a+x\cdot0=a[/tex]
times 0 would be zero,right?
 
52
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never mind, sorry.
 
52
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lets just go back to the original equation
[tex]fsingle-quote(x)=e^{g(x)}gsingle-quote(x)[/tex]
what is g'(X) equall to
 
1,750
1
read my previous posts!!!
 
52
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g'= -m
Solved.
thanks for the help
 
1,750
1
:) anytime
 

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