What is (\partial)/(\partial x)(4cos(xy^2)) ?

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

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(\partial)/(\partial x)(4cos(xy^2))

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

\int (x^{3} - \frac{1}{2 x} + 1) d x

w \to - 1 + lim (w + ∣ 3 w + 3 ∣)

\int - sin (- x) d x

2 x^{2} y^{^{''}} + 6 x y^{^{'}} + 2 y = 0

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

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