34x^12 - 3x^5 + 23xy^3 - y^2 - xy
If you can answer it (correctly), you get a cookie!
34x^12 - 3x^5 + 23xy^3 - y^2 - xy
If you can answer it (correctly), you get a cookie!
I have no idea if I understand your question right since you got two unkown variables in the same function, and we don't really deal with that (so far) at the uni. But sure, I'll have a shot at that;
D(34x^12) is obv 34*12x^11 = 408x^11
D(3x^5) is also obv 3*5x^4 = 15x^4
23xy^3, here is a little trickier one (albeit still easy for me if I understand your question right). if we are supposed to derive both x and y (which I don't really know if we are, your first question doesn't tell me as much) you got a specific rule as to how that works, I have no idea what it's called in English though. But basically f(x)*f(y) (or normally f(x)*g(x)) would have the derivative f'(x)*f(y) + f(x)*f'(y). This would result in 23*y^3 + 23x*3y^2, which is equal to 23y^2(y + 3x).
D(y^2) = 2y
D(xy) = y + x (using the same rule as mentioned above.)
Now you got:
408x^11 - 15x^4 + 23y^2(y + 3x) - 2y - (y + x) = 408x^11 - 15x^4 + 23y^2(y + 3x) - 2y - y - x
Bunched together:
408x^11 - 15x^4 + 23y^2(y + 3x) - 3y - x, which could be written a bit differently but not necessarily easier. Your teacher should be able to accept that solution, if that is what it's all about.
Note, I did not use any program like mathlab or wolframalpha for this. Pure head calculus.
However I'm still not sure if I understood your question correctly. Might be that this is meant to be some kind of special case with two unknown variables, and in that case I'm probably not gonna be of much use since I haven't gotten there yet.
I don't do calculus for anything less than double chocolate chip.
And it better be warm, fresh and gooey.
Putin khuliyo
(d)/(dx)(-3 x^5+34 x^12-x y-y^2+23 x y^3) = 408 x^11-15 x^4+23 y^3-y ?
assuming it's a function f with x and y being variables:
d(f)/d(x)=408x^11-15x^4+23y^3-y
d(f)/d(y)=69xy^2-2y-x