What is (\partial)/(\partial y)(sin^4(1-4xy+y^2)) ?

The answer to (\partial)/(\partial y)(sin^4(1-4xy+y^2)) is 4sin^3(1-4xy+y^2)cos(1-4xy+y^2)(2y-4x)

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(\partial)/(\partial y)(sin^4(1-4xy+y^2))

\frac{\partial}{\partial y} (sin^{4} (1 - 4 x y + y^{2}))

4 sin^{3} (1 - 4 x y + y^{2}) cos (1 - 4 x y + y^{2}) (2 y - 4 x)

Solution steps

\frac{\partial}{\partial y} (sin^{4} (1 - 4 x y + y^{2}))

Treat

x

as a constant

Apply the chain rule:

4 (sin (1 - 4 x y + y^{2}))^{3} \frac{\partial}{\partial y} (sin (1 - 4 x y + y^{2}))

= 4 (sin (1 - 4 x y + y^{2}))^{3} \frac{\partial}{\partial y} (sin (1 - 4 x y + y^{2}))

\frac{\partial}{\partial y} (sin (1 - 4 x y + y^{2})) = cos (1 - 4 x y + y^{2}) (2 y - 4 x)

= 4 (sin (1 - 4 x y + y^{2}))^{3} cos (1 - 4 x y + y^{2}) (2 y - 4 x)

Simplify

= 4 sin^{3} (1 - 4 x y + y^{2}) cos (1 - 4 x y + y^{2}) (2 y - 4 x)

f (x) = 500 - y - \frac{4 y}{y + 3}

\int \frac{1}{4 x ^{2} + 5} d x

x \to 0 lim (cos (π) x)

\int 3 x^{- \frac{2}{3}} d x

\int e^{2 x} (sin (3 x)) d x

What is (\partial)/(\partial y)(sin^4(1-4xy+y^2)) ?
The answer to (\partial)/(\partial y)(sin^4(1-4xy+y^2)) is 4sin^3(1-4xy+y^2)cos(1-4xy+y^2)(2y-4x)