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.
can you explain how you find the partial derivative with respect to x? Short sentance...
you basically do derivative just with respect to x and treating y as a constant
So Mproof, Hmm If you take partials, you'll be treating the OTHER variable as a constant, meaning you won't have the product rule as it might seem at first glance. Does that help? :O
like 3x^2y+2 fx=6xy and Fy is 3x^2
right. So what is confusing you about this problem?
would Hx be 0?
no. What is the regular derivative of f(x) = e^x?
Think of the equation as Ce^x when taking the partial WRT x. Maybe that will help :)
but don't you have to take a derivative of the (y^2+1) with respect to x?
no, thats just a constant attached to e^x :d
Maybe one thing you can do to convince yourself is, distribute the e^x to each term in the brackets. Then think about what you have :o
so the derivative with respect to x will be the same as the given problem
ya :) good
ooo I get it
with respect to Y would it be 2ye^x?
(y^2 + 1)e^x = y^2 e^x + e^x Hy = 2y e^x + 0 Yes, very good ^^