## IsTim Group Title Express each function as a power with a rational exponent, and then differentiate. Express each answer using positive exponents. y=sqrt(2x-3x^5) 2 years ago 2 years ago

1. IsTim Group Title

Well. I'm just doing the first part. I got 1/2(2-15x^4)^-1/2

2. dumbcow Group Title

ok sqrt is same as exponent of 1/2 then apply chain rule to differentiate good..wait don't change inside...multiply whole thing by derivative of inside

3. dumbcow Group Title

$(2-15x^{4})\frac{1}{2}(2x-5x^{3})^{-1/2}$

4. dumbcow Group Title

**typo 3x^5

5. IsTim Group Title

In my textbook, it says: $y=(2x-3x^5)^{1/2}$

6. dumbcow Group Title

?? yes that is original function right

7. IsTim Group Title

No, the answer. Well, the first half that isn't dy/dx.

8. dumbcow Group Title

oh yeah...sqrt is same as exponent of 1/2

9. IsTim Group Title

But I don't know how to get $dy/dx=(2=15x^4 )/(2\sqrt{2x-3x^5}$

10. IsTim Group Title

Also, why is their exponent positive, but ours is negative?

11. dumbcow Group Title

same answer i posted....ours is negative because its in numerator, when you flip it to denominator the sign of exponent changes

12. IsTim Group Title

Ok. I understand the first part. Thanks. I'll call out if I need help on the second part.

13. IsTim Group Title

Um, to find the derviative I use 1/2(2x−5x^3)−1/2 right.

14. dumbcow Group Title

whoah ...um we were doing 2nd part (finding derivative this whole time) or at least thats what i was doing

15. IsTim Group Title

Oh no! How do I do the first part?

16. dumbcow Group Title

sqrt is same as exponent of 1/2

17. IsTim Group Title

To get dy/dx=(2=15x^4)

18. IsTim Group Title

/(22x−3x5−−−−−−−√

19. IsTim Group Title

I guess I can't copy Latex....

20. IsTim Group Title

dy/dx=(2=15x^4)/(2sqrt(2x−3x^5))

21. IsTim Group Title

How do I figure that out?

22. dumbcow Group Title

let me try this, part 1 is super easy all you are doing is rewriting it using rational exponents $\large \sqrt[n]{x} = x^{1/n}$ $\sqrt{2x-3x^{5}} = (2x-3x^{5})^{1/2}$ part 2: find derivative using chain rule u = 2x-3x^5 du = 2-15x^4 $\frac{d}{du} u^{1/2} = \frac{1}{2}u^{-1/2} = \frac{1}{2\sqrt{u}}$ multiply by du and substitute 2x-3x^5 back in for u $\rightarrow \frac{2-15x^{4}}{2\sqrt{2x-3x^{5}}}$

23. IsTim Group Title

How'd you get du = 2-15x^4?

24. IsTim Group Title

@dumbcow When you have time...

25. dumbcow Group Title

isn't that what you got in your very 1st post ^^ ?? anyway, its the derivative of 2x-3x^5 using power rule $\large \frac{d}{dx} x^{n} = n*x^{n-1}$

26. IsTim Group Title

Ok. thanks.