What is (\partial)/(\partial x)(cos(xy)+e^{yz}+ln(zx)) ?

The answer to (\partial)/(\partial x)(cos(xy)+e^{yz}+ln(zx)) is 1/x-ysin(xy)

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(\partial)/(\partial x)(cos(xy)+e^{yz}+ln(zx))

\frac{\partial}{\partial x} (cos (x y) + e^{yz} + ln (z x))

\frac{1}{x} - y sin (x y)

Solution steps

\frac{\partial}{\partial x} (cos (x y) + e^{yz} + ln (z x))

Treat

y, z

as constants

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (cos (x y)) + \frac{\partial}{\partial x} (e^{yz}) + \frac{\partial}{\partial x} (ln (z x))

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

\frac{\partial}{\partial x} (e^{yz}) = 0

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

= - sin (x y) y + 0 + \frac{1}{x}

Simplify

= \frac{1}{x} - y sin (x y)

f^{9}

\frac{d y}{d x} = \frac{( 2 x + 1 )}{3 y}

\frac{d y}{d x} = 2 y - 4 x

f (x) = x^{2} (6 x + 1)^{3}

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

What is (\partial)/(\partial x)(cos(xy)+e^{yz}+ln(zx)) ?
The answer to (\partial)/(\partial x)(cos(xy)+e^{yz}+ln(zx)) is 1/x-ysin(xy)