XY sin (x + y) = 1 find the derivative of Y

Derivative (XY) '= x' * y + X * y '= y + X * y' [sin (x + y)] '= cos (x + y) * (x + y)' = (1 + y ') cos (x + y) = cos (x + y) + cos (x + y) * y'1' = 0, so y + X * y '- cos (x + y) - cos (x + y) * y' = 0y '= [y-cos (x + y)] / [cos (x + y) - x]

Find the derivative of y to the third power of X + XY sin (π y) = o I hope the sooner the better. I'm in urgent need

Take the derivative of both sides of the equation with respect to y
X-πcos(πY)=O
therefore
cos(πY)=X/π
Y={arccos(X/π)}/π
Derivative of Y
Y '= - 1 / π * {1 / under the root sign [1 - (x / π) ^ 2]}

Find the derivative y ′ X of the function y = f (x) determined by the Y square of equation E - x square of equation E + xy = 0

e^y-e^x+xy=0
e^y*y’-e^x+y+xy'=0
y'=(e^x-y)/(e^y+x)

The derivative of y = sin ^ 3 (1 / x) is

y'
=(sin^3(1/x))'
=3sin^2(1/x)*(sin(1/x))'
=3sin^2(1/x)*cos(1/x)*(1/x)'
=-(3sin^2(1/x)cos(1/x))/x^2

Sin ^ 2 x derivative Why is sin squared x derivative not cos squared x?

(sin^2(x))'=2sin(x)(sin(x))'=2sin(x)cos(x)

Find the n-order derivative of y = (E Λ x) SiNx, and the answer is y (n) = e Λ x (SiNx + sin (x + π / 2) +... + s Find the n-order derivative of y = (E Λ x) SiNx The answer is y (n) = e Λ x (SiNx + sin (x + π / 2) +... + sin (x + n π / 2)) =e∧x（（sinx＋sin（x＋nπ/2））+（sin（x＋π/2）＋sin(x＋（n-1）π/2）)+…） =2∧（n/2） e∧x sin（x＋nπ/4） How did the last step come about

The first step of your answer is wrong, but the final answer is right. You make n = 2 and try. Is it right after the first equal sign?
The first is the wrong solution, and the last is the correct answer. Of course, I don't know how the last step came
In other words, why do you do it in such a tangled way? If there is no restricted solution to the problem, I think Euler formula is more convenient:

XY sin (x + y) = 1 find the derivative of Y