Why is the derivative equation of X & # 178; + XY + Y & # 178; = 4 more than 2x + y + XY '+ 2yy'? X & # 178; derivative is 2x, XY, derivative y, Y & # 178; derivative 2Y, I don't know why there is more y, and it was originally 2Y, actually 2yy·

Why is the derivative equation of X & # 178; + XY + Y & # 178; = 4 more than 2x + y + XY '+ 2yy'? X & # 178; derivative is 2x, XY, derivative y, Y & # 178; derivative 2Y, I don't know why there is more y, and it was originally 2Y, actually 2yy·

Because y is also a function of X
So XY can be seen as X * y
Multiplication formula by derivation
(x*y)'=x'*y+x*y'=y+xy'

Given that the curve y = ax + B / X is tangent to y = x ^ 3 at point (1,1), find the value of a and B

Dy / DX (derivative of Y function) = A-B / x ^ 2, x = 1, the derivative is A-B, y = x ^ 3 is 3x ^ 2, x = 1, the derivative is 3, that is, A-B = 3, then substitute the point (1,1) into the first function, a + B = 1, a = 2, B = - 1

If we know a-3b-178; + m + 9b-178; = (a-3b-178;), then M =?; if x + 2Y = 6, xy = 4, then x-178; + 4y-178; =?

(a-3b)²
=a²-6ab+9b²
So m = - 6ab
x+2y=6
Square on both sides
x²+4xy+4y²=36
x²+4xy
=36-4xy
=36-16
=20

Let f (x) = vector a × (vector B + vector C), where vector a = (SiNx)
Let f (x) = vector a * (vector B + vector C), where vector a = (SiNx, - cosx), vector b = (SiNx, - 3cosx), vector C = (- cosx, SiNx), X ∈ R
The image of the function y = f (x) is translated according to the vector D, so that the translated image is centrosymmetric with respect to the coordinate origin, and the d with the smallest length is obtained

Let D (h, K) be a vector
So x '= X-H; y' = y-k
x=x’-h ； y=y’-k
Then we take the above formula into the original f (x)
Y '- H = 2 + √ 2Sin (2x-2h-3 π / 4)
Now see the condition of "make the image obtained after translation be centrosymmetric with respect to the coordinate origin"
That is to say, when x = 0, the equation g (0) = 0
So - 2h-3 π / 4 = k π
H = 3 π / 8-K π / 2
Then we get D (3 π / 8-K π / 2, - 2)
The key to solve this problem is to set the vector according to the translation method
This x '= X-H; y' = y-k
x=x’-h ； y=y’-k
F (x) = vector a * (B + C)
From the problem f (x) = (SiNx, - cosx) * (SiNx cosx, - 3cosx + SiNx)
f(x)=sinx(sinx-cosx)-cosx(-3cosx+sinx)
=sinxsinx-sinxcosx+3cosxcosx-sinxcosx
=sinxsins+3cosxcosx-2sinxcosx
=sinxsinx+cosxcosx+2cosxcosx-2sinxcosx
=cos2x-sin2x
=Radical 2 / 2 sin (2x + 45 degrees)

Given x ^ n = 5, y ^ n = 3, find the value of {x ^ 2Y} ^ 2n

=X^4n*y^2n
=(X^n)^4*(y^n)^2
=5^4*3^2
=625*9
=5625

Let's ask the function f (x) = xsin1 / x, x > 0, f (x) = 10, x = 0, 5 + X05, x = 0 at x = 0

When x tends to zero, the limit is zero
When x tends to zero and 1 / X tends to infinity or infinitesimal, then sin (1 / x) is a bounded function between - 1 and 1
Then xsin (1 / x) is infinitesimal times bounded function, or infinitesimal, that is, the limit is 0

Given xn = 5, yn = 3, find: (1) (X2Y) 2n; (2) x3n △ y4n

（1）∵xn=5，yn=3，∴（x2y）2n=x4ny2n=（xn）4（yn）2=54×32=5625；（2）∵xn=5，yn=3，∴x3n÷y4n=（xn）3÷（yn）4=53÷34=12581．

Piecewise function f (x) = xsin1 / x, x > 0, a + x ^ 2, X

If x + y = 3M, X-Y = N3, then X4 + y4-2x2y2=______ ．

∵ x + y = 3M, X-Y = N3, ∵ X4 + y4-2x2y2 = (x2-y2) 2 = (X-Y) 2 (x + y) 2 = n29 × (3m) 2 = n2m2

The limit X of F (x) = xsin1 / X tends to 0

f(x)=xsin(1/x)；
Because - 1 ≤ sin (1 / x) ≤ 1;
So - x ≤ f (x) ≤ X;
lim(-x)=0,lim(x)=0；
According to the pinch principle, limf (x) = 0 when x tends to 0;