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

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

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(\partial)/(\partial y)(sin(x^2+xy))

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

cos (x^{2} + x y) x

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

= cos (x^{2} + x y) x

y^{^{''}} - y = 18 t

\frac{d}{d x} (- x^{2} + 6 x)

f (x) = - 5 x^{2}, at x = 6

\frac{\partial}{\partial x} (x - y)

\frac{d}{d x} (2 x e^{6 x} + e^{6 x} \cdot 6 x^{2})

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