First derivative: chain rule: easy for you guys

[physicsed]

[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

 

[rocomath]

$$f(x)=e^{g(x)}$$

$$f'(x)=e^{g(x)}g'(x)$$

$$y=e^{-mx}$$

$$y'=-me^{-mx}$$

 

[physicsed]

yes!

 

[rocomath]

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

 

[physicsed]

how did you get g'(x)
i thought the (m) was constant thus making the the derivative equal to 0

 

[rocomath]

Do this one for me ...

$$y=ax$$

What is it's derivative?

 

[physicsed]

gosh!
well i would use the product rule or maybe not.
f= x g=ax
f'= 1 g'=product rule
high school calc.

 

[rocomath]

Don't worry about the product rule when you're differentiating with a constant.

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

$$\frac{d}{dx}=cf'(x)$$

$$y=ax$$

$$y'=a$$ by constant rule

$$y'=a+x\cdot0=a$$ by product rule

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

 

[physicsed]

$$ysingle-quote=a+x\cdot0=a$$
times 0 would be zero,right?

 

[physicsed]

never mind, sorry.

 

[physicsed]

lets just go back to the original equation
$$fsingle-quote(x)=e^{g(x)}gsingle-quote(x)$$
what is g'(X) equall to

 

[rocomath]

read my previous posts!

 

[physicsed]

g'= -m
Solved.
thanks for the help

 

[rocomath]

:) anytime