What is (\partial)/(\partial x)(9y^4cos(x^2)) ?

The answer to (\partial)/(\partial x)(9y^4cos(x^2)) is -18y^4xsin(x^2)

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(\partial)/(\partial x)(9y^4cos(x^2))

\frac{\partial}{\partial x} (9 y^{4} cos (x^{2}))

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

Solution steps

\frac{\partial}{\partial x} (9 y^{4} cos (x^{2}))

Treat

y

as a constant

Take the constant out:

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

= 9 y^{4} \frac{\partial}{\partial x} (cos (x^{2}))

Apply the chain rule:

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

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

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

= 9 y^{4} (- sin (x^{2}) \cdot 2 x)

Simplify

= - 18 y^{4} x sin (x^{2})

d er i v a t i v e f (x) = \frac{1}{x ^{2} - 9}

in v erse l a pl a ce \frac{1}{2}

\int x^{2} \cdot e^{- 3 x} d x

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

\int \frac{14}{2 x + x x} d x

What is (\partial)/(\partial x)(9y^4cos(x^2)) ?
The answer to (\partial)/(\partial x)(9y^4cos(x^2)) is -18y^4xsin(x^2)