## cunninnc 2 years ago find the derivative of y=log10(5x^4-2/x^3)

1. MrMoose

this involves both the chain rule and the quotient rule

2. Mimi_x3

or you can use log laws to change it first; then diferentiate it, would be easier.

3. cunninnc

@failmathmajor only the 2/x^3 is a fraction

4. Mimi_x3

is it $\log\left(5x^{4}-\frac{2}{x^{3}}\right) ?$

5. MrMoose

that would be much easier

6. MrMoose

so using log properties you can say that $\log_{10} {x} = \ln x / \ln 10$

7. MrMoose

ln10 is a constant, you can factor that out

8. MrMoose

can you use the chain rule fluently?

9. cunninnc

y'=pu(x) * u'(x) right?

10. MrMoose

d(f(g(x))/dx = (dg(x)/dx)*(df(g(x))/dx) is the way I know it

11. Mimi_x3

Or..$y= \log\left(5x^{4}-\frac{2}{x^{3}}\right) y = \log\left(\frac{5x^{7}-2}{x^{3}}\right) => y = \log\left(5x^{7}-2\right) - \log(x^3)$ Might be easier; than the quotient role and chain rule..

12. MrMoose

that may look confusing now that I see it

13. MrMoose

don't need the quotient in this case: can treat /x^3 as a x^-3

14. Mimi_x3

@cunninnc: are you able to do it now?

15. MrMoose

and you would still need the chain rule for that anyway

16. MrMoose

but enough arguing

17. Mimi_x3

Did an argument started? lol

18. MrMoose

$(d(\ln (5x^4 - 2x^{-3}))/dx )/ \ln10$ is what it simplifies down to, to be concise

19. cunninnc

@Mimi_x3 kinda .... i see mr. moose got 2x^-3 where does ^-3 come froms

20. MrMoose

$2/x^3= 2 * x^{-3}$

21. Mimi_x3

Sorry i don't know what MrMoose is doing. @MrMoose: Use \frac{x}{y} for fractions :)

22. Mimi_x3

Or why not try the method that i used :) $\frac{d}{dx} \log\left(5x^{7}-2\right) -\frac{d}{dx} \log(x^3)$

23. MrMoose

when you divide you subtract exponents, so that is equivalent to saying $2 * \frac{x^0}{x^3}$ then subtract exponents in division

24. MrMoose

$\frac{\frac{d(\ln(5x^4−2x^{−3}))}{dx}}{\ln10}$

25. MrMoose

@catamountz15 what?

26. MrMoose

I am almost entirely sure that that isn't a form of the answer.

27. catamountz15

Here are the steps in to solving this problem.

28. MrMoose

that isn't what you wrote though

29. catamountz15

My apologies that was an answer to a different problem.