What is (\partial)/(\partial x)(cos(x^2)*sin(z*y)) ?

The answer to (\partial)/(\partial x)(cos(x^2)*sin(z*y)) is -2xsin(zy)sin(x^2)

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(\partial)/(\partial x)(cos(x^2)sin(zy))

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

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

Solution steps

\frac{\partial}{\partial x} (cos (x^{2}) sin (zy))

Treat

z, y

as constants

Take the constant out:

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

= sin (zy) \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

= sin (zy) (- sin (x^{2}) \cdot 2 x)

Simplify

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

\frac{\partial}{\partial x} (17 - 2 x^{2} - y^{2})

\frac{\partial}{\partial x} (3 x^{2} y^{2} - 4)

x^{3} + 4 x

y = 7 cos (2 x), y = 7 - 7 cos (2 x)

\frac{d}{d x} (5 x^{2})

What is (\partial)/(\partial x)(cos(x^2)*sin(z*y)) ?
The answer to (\partial)/(\partial x)(cos(x^2)*sin(z*y)) is -2xsin(zy)sin(x^2)