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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

t \to + 0 lim (\frac{e ^{t} - 1}{t ^{9}})

f (x) = \frac{3}{x}, (6, \frac{1}{2})

\int 32 sin^{4} (2 x) d x

\frac{d}{d y} ((x^{2} + y^{2}) arctan (\frac{y}{x}))

e^{x}, x = 2

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