Let the equation y-cos (x ^ 2 + y ^ 2) = x of function y = g (x), find dy / DX

Let the equation y-cos (x ^ 2 + y ^ 2) = x of function y = g (x), find dy / DX

Answer:
y-cos(x^2+y^2)=x
Derivation of X on both sides: y '+ sin (x ^ 2 + y ^ 2) * (2x + 2yy') = 1
[2ysin(x^2+y^2)+1]*y'=1-2xsin(x^2+y^2)
y'=[1-2xsin(x^2+y^2)]/[2ysin(x^2+y^2)+1]
So:
dy/dx=[1-2xsin(x^2+y^2)]/[2ysin(x^2+y^2)+1]

Let y = y (x) be a function determined by the equation cos (x + y) + y = 1, and find the derivative dy / DX

Cos (x + y) + y = 1, both sides take the derivative of X at the same time
-(1+y~)sin(x+y)+y~=0
Available:
=(1+y~)sin(x+y)
=sin(x+y)/(1-sin(x+y))

Let the equation ∫ (y, 0) e ^ (T ^ 2) DT + ∫ (1, x ^ 2) cos √ TDT determine the function of Y as X, and find dy / DX (y, 0) means that y is the upper limit of the integral and 0 is the lower limit of the integral. (1, x ^ 2) the same

The title formula is missing. There is no equal sign. It is not a function, but an algebraic formula

Let the function y = y (x) be determined by the equation x ^ 2 + y ^ 2 = 1, and find dy / DX

Derivation of X on both sides
2x + 2y * dy/dx=0
dy/dx = -x/y
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Let the function y = YF (x) be determined by the equation x + cos (x + y) = 0 in [0, Pai], then dy / DX x = 0= a -1 b 0 c pai/2 d 2

B
Take the differential on both sides of the equation x + cos (x + y) = 0, and get DX - sin (x + y) d (x + y) = 0
That is, DX - sin (x + y) DX + sin (x + y) dy = 0, and [1 - sin (x + y)] DX = - sin (x + y0dy
Thus dy / DX = | [1 - sin (x + y)] / sin (x + y) | (*)
When x = 0, y = Pai / 2 is obtained by substituting into the original equation,
Then substitute the obtained y = Pai / 2 and x = 0 into the (*) formula to get dy / DX x = 0 = 0, and choose B

XY sin (π y ^ 2) = 0 find dy / DX

y+xy'-cos(πy ²) 2πyy'=0
y=[2πycos(πy ²)- x]y'
y'=y/[2πycos(πy ²)- x]
Namely: dy / DX = Y / [2 π ycos (π y) ²)- x]