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.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

Camila have we met before ?

No I don't think so. I would first subtract 1 from the xy and put it in front of the e correct?

Where are you from?

How would you derive the e^xy though? Because the product rule is f'(x)g(x)+f(x)g'(x)

I think it would be 3e^x^2?

\[\Large\rm \left(e^{x^3}\right)'=e^{x^3}(x^3)'\]\[\Large\rm \left(e^{x^3}\right)'=e^{x^3}(3x^2)\]

Would you have to multiply it by its inverse?

What do you think lady Cam? :d
too confusing?

I am very confused.

I know how to do the chain rule when it doesn't involve natural logs but now it got confusing.

But, if working with logs is really confusing, we can maybe avoid taking that route.

Woops, your derivative looks a little wonky.

It equals 0

good good good, it's just a fancy looking constant. but it's still a constant.
so derivative is 0 :)

\[\Large\rm y+xy'=0\]So that finishes up the differentiation process at least, ya? :)

Now plug and chug

So its ln4+0=0 so it would just be ln4?

you're plugging in `1 for x`, and `ln4 for y`,
i'm not sure where you're getting that 0 from :o

Because you said xy'

you're trying to solve for y',
you don't plug anything into that.

Ooooooo okay so it would be ln4+1=0 and then when you solve it would be -ln4+4?

\[\Large\rm y'=-\ln4\]Yayyyy good job \c:/

\[\Large\rm \frac{dy}{dx}=-\ln 4\]

Thank you so much for helping me through this!!

Oh okay