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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x, g = cos (2 + x^{2} y)

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

\frac{\partial}{\partial x} (x) = 1

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

= y^{2} (1 \cdot cos (2 + x^{2} y) + (- sin (2 + x^{2} y) \cdot 2 y x) x)

Simplify

y^{2} (1 \cdot cos (2 + x^{2} y) + (- sin (2 + x^{2} y) \cdot 2 y x) x) : y^{2} (cos (2 + x^{2} y) - 2 x^{2} y sin (2 + x^{2} y))

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

2 sin (x), [0, π]

f (x) = \frac{5}{2 5 x + 6}, at x = 6

\frac{\partial}{\partial x} (3 x)

y^{^{'}} - 4 y = 7 e^{4 x}

\int \frac{1}{x ^{2} + 11 x + 30} d x

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