Homework Help: First derivative: chain rule: easy for you guys

1. Mar 13, 2008

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

2. Mar 13, 2008

rocomath

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

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

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

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

Last edited: Mar 13, 2008
3. Mar 13, 2008

physicsed

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

4. Mar 13, 2008

rocomath

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

5. Mar 13, 2008

physicsed

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

6. Mar 13, 2008

rocomath

Do this one for me ...

$$y=ax$$

What is it's derivative?

7. Mar 13, 2008

physicsed

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

8. Mar 13, 2008

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?

9. Mar 13, 2008

physicsed

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

10. Mar 13, 2008

physicsed

never mind, sorry.

11. Mar 13, 2008

physicsed

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

12. Mar 13, 2008

rocomath

13. Mar 13, 2008

physicsed

g'= -m
Solved.
thanks for the help

14. Mar 13, 2008

:) anytime