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

1. anonymous

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. anonymous

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

4. Mimi_x3

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

5. anonymous

that would be much easier

6. anonymous

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

7. anonymous

ln10 is a constant, you can factor that out

8. anonymous

can you use the chain rule fluently?

9. anonymous

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

10. anonymous

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. anonymous

that may look confusing now that I see it

13. anonymous

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. anonymous

and you would still need the chain rule for that anyway

16. anonymous

but enough arguing

17. Mimi_x3

Did an argument started? lol

18. anonymous

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

19. anonymous

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

20. anonymous

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

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. anonymous

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

24. anonymous

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

25. anonymous

@catamountz15 what?

26. anonymous

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

27. anonymous

Here are the steps in to solving this problem.

28. anonymous

that isn't what you wrote though

29. anonymous

My apologies that was an answer to a different problem.