What is (\partial)/(\partial x)(3zcos(4xy)+x-7y) ?

The answer to (\partial)/(\partial x)(3zcos(4xy)+x-7y) is -12zysin(4xy)+1

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(\partial)/(\partial x)(3zcos(4xy)+x-7y)

\frac{\partial}{\partial x} (3 z cos (4 x y) + x - 7 y)

- 12 zy sin (4 x y) + 1

Solution steps

\frac{\partial}{\partial x} (3 z cos (4 x y) + x - 7 y)

Treat

z, y

as constants

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{\partial}{\partial x} (3 z cos (4 x y)) + \frac{\partial}{\partial x} (x) - \frac{\partial}{\partial x} (7 y)

\frac{\partial}{\partial x} (3 z cos (4 x y)) = - 12 zy sin (4 y x)

\frac{\partial}{\partial x} (x) = 1

\frac{\partial}{\partial x} (7 y) = 0

= - 12 zy sin (4 y x) + 1 - 0

Simplify

= - 12 zy sin (4 x y) + 1

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

f (x) = \frac{2}{x - 6}, at x = 8

\int_{2}^{5} \frac{x}{x ^{2} + 4 x + 20} d x

(x^{2} - 16 x + 64) (x^{2} + 64 x + 4)

\int \frac{sec ( \frac{1}{x} )}{x ^{2}} d x

What is (\partial)/(\partial x)(3zcos(4xy)+x-7y) ?
The answer to (\partial)/(\partial x)(3zcos(4xy)+x-7y) is -12zysin(4xy)+1