Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

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

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

Well. I'm just doing the first part. I got 1/2(2-15x^4)^-1/2
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
\[(2-15x^{4})\frac{1}{2}(2x-5x^{3})^{-1/2}\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

**typo 3x^5
In my textbook, it says: \[y=(2x-3x^5)^{1/2}\]
?? yes that is original function right
No, the answer. Well, the first half that isn't dy/dx.
oh yeah...sqrt is same as exponent of 1/2
But I don't know how to get \[dy/dx=(2=15x^4 )/(2\sqrt{2x-3x^5}\]
Also, why is their exponent positive, but ours is negative?
same answer i posted....ours is negative because its in numerator, when you flip it to denominator the sign of exponent changes
Ok. I understand the first part. Thanks. I'll call out if I need help on the second part.
Um, to find the derviative I use 1/2(2x−5x^3)−1/2 right.
whoah ...um we were doing 2nd part (finding derivative this whole time) or at least thats what i was doing
Oh no! How do I do the first part?
sqrt is same as exponent of 1/2
To get dy/dx=(2=15x^4)
/(22x−3x5−−−−−−−√
I guess I can't copy Latex....
dy/dx=(2=15x^4)/(2sqrt(2x−3x^5))
How do I figure that out?
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 \[\rightarrow \frac{2-15x^{4}}{2\sqrt{2x-3x^{5}}}\]
How'd you get du = 2-15x^4?
@dumbcow When you have time...
isn't that what you got in your very 1st post ^^ ?? anyway, its the derivative of 2x-3x^5 using power rule \[\large \frac{d}{dx} x^{n} = n*x^{n-1}\]
Ok. thanks.

Not the answer you are looking for?

Search for more explanations.

Ask your own question