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

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

Popular >

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

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

cos (x^{2} y - 4 x y^{2}) (2 y x - 4 y^{2})

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

cos (x^{2} y - 4 x y^{2}) \frac{\partial}{\partial x} (x^{2} y - 4 x y^{2})

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

\frac{\partial}{\partial x} (x^{2} y - 4 x y^{2}) = 2 y x - 4 y^{2}

= cos (x^{2} y - 4 x y^{2}) (2 y x - 4 y^{2})

\int \frac{7}{( t ^{2} + 1 ) ^{\frac{3}{2}}} d t

\frac{4}{3 x ^{2}}

f (x) = x^{2} - \frac{1}{x}, (- 1, 0)

\int_{0}^{2} \frac{x + 2}{x - 4} d x

x^{\frac{2}{3}}

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