First simplify and then evaluate the 3rd power of X * (- Y's 3rd power) square + (- 1 / 3xy's Square), where x = 1 / 2, y = - 1

First simplify and then evaluate the 3rd power of X * (- Y's 3rd power) square + (- 1 / 3xy's Square), where x = 1 / 2, y = - 1

Original formula = x ³ y^6-1/27x ³ y^6
=26/27x ³ y^6
=26/27*1/8*1
=13/108

The result of calculating 2x • (- 3xy) 2 • (- X2Y) 3 is __

2x•（-3xy）2•（-x2y）3，
=2x•9x2y2•（-x6y3），
=-18x9y5．

X Cubic + 3xy + y cubic - 1

Don't bother to copy the process. You can push it back
(x + y - 1)*(x^2 - x*y + x + y^2 + y + 1)

Let the function z = x2y2, where x = Sint, y = cost, find DZ / dt 2 is the power of two

This problem belongs to the derivation category of composite function!
The detailed process is as follows:
dz/dt
=(x^2)'y^2+x^2(y^2)'
=2sint(cost)^3-(sint)^3*2cost
=2sintcost[(cost)^2-(sint)^2]
=sin2tcos2t
=(1/2)*sin4t

A mathematical problem about calculus If D / DX (sin ^ 3 2x) = 6sin ^ 2 2x cos2x, Calculate the value of COS ^ 3 2x DX. (range: between 0 and π / 4) Thank you~

The value with DX is an infinitesimal quantity and is not computable

Ask for a math problem in calculus When n tends to ∞, find the limit of ∑ {n / [(n + k) square]}. How is k calculated from 1 to n?

Your writing is really nonstandard. I read it wrong. Wait... First. 1 / (n + 1) ^ 2 > 1 / (n + 1) (n + 2) = 1 / (n + 1) - 1 / (n + 2) all add up, the original formula > n * (1 / (n + 1) - 1 / (2n + 1)) = n ^ 2 / (2n ^ 2 + 3N + 1) 1 / (n + 1) ^ 2 < 1 / N (n + 1) = 1 / N - 1 / (n + 1) the poles of the above two formulas