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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

= 2 \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}

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

Simplify

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

\frac{( 4 x ^{2} + 8 x + 6 )}{x}

x \to 0 lim (∣ x - 4 ∣)

\int_{- 3}^{1} (x^{2} - 4 x) d x

\frac{d}{d x} (ln (ln (x^{2} + 1)))

y^{^{''}} + a^{2} y = 0

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