## amyna one year ago diffrentiate: g(x)=4^3x^2 Thank you for your help!

1. amyna

|dw:1441504840472:dw|

2. triciaal

differentiate power raised to a power think we can do substitution

3. triciaal

|dw:1441505133978:dw|

4. Astrophysics

Try logarithms

5. triciaal

maybe the log function

6. triciaal

right on time huh

7. Astrophysics

Haha :P

8. amyna

lol i dont understand. what you wrote, is that the correct way to do this problem or not?

9. beginnersmind

Nah. Start by rewriting |dw:1441505728153:dw|

10. amyna

oh okay so you have to use logs and rewrite it. is this true for all problems that may look similar to this?

11. beginnersmind

Generally base e is easier to deal with. And changing bases only changes the exponent by a constant factor. So it almost always helps.

12. amyna

ok thanks!

13. amyna

wait then how do i solve it after rewriting it? lol i forgot how to do that part!

14. beginnersmind

Use the chain rule.

15. amyna

ok thanks. i think i got it from here

16. amyna

no i don't get it. i don't know how to solve it. i tried using the chain rule

17. beginnersmind

Hm, you might need to review the chain rule then. I can go through this example if you want but I don't think it will be much help in general.

18. beginnersmind

@amyna

19. amyna

yes please! i would greatly appreciate that! :)

20. freckles

$y=(f(x))^{g(x)} , \text{ assume } f(x)>0 \\ \\ \text{ take } \ln( ) \text{ of both sides } \\ \ln(y)=\ln((f(x))^{g(x)}) \\ \ln(y)=g(x) \ln(f(x)) \text{ by use of power rule for logarithms } \\ \\ \text{ now differentiate both sides } \\ \frac{y'}{y}=g'(x) \cdot \ln(f(x))+g(x) \cdot \frac{f'(x)}{f(x)} \\ \text{ by a whole bunch of rules :p } \\ \text{ left hand side I just used chain rule } \\ \text{ right hand side I used product rule and chain rule }$ $y'=\{g'(x) \ln(f(x))+g(x) \frac{f'(x)}{f(x)} \} y \\ \text{ note: this step I just multiplied both sides by } y$ $\text{ now remember } y=(f(x))^{g(x)} \\ \text{ so make this replacement and you are done} \\$

21. beginnersmind

Ok, I'll try to explain how to apply the chain rule. You might want to look at this lesson from MIT OCW as well: http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/1.-differentiation/part-a-definition-and-basic-rules/session-11-chain-rule/ You look at $$e^{(ln4)*3*x^2}$$. If it was e^x the derivative would be e^x. But now the exponent is a function of x. So the result is e to that function of x _multiplied by the derivative of that function_. So $$[e^{g(x)}]' = g'(x)* e^{g(x)}$$ In this case g(x) = (ln4)*3*x^2 g'(x) = (ln4)*6*x so the final result is $(ln4)*6x*e^{ln4*3*x^2}$ which you can rewrite as $(ln4)*6x*4^{ln4*3*x^2}$ using the same idea with logarithms that we started with.