What is (\partial)/(\partial x)(ln(x+y)+x*cos(y)) ?

The answer to (\partial)/(\partial x)(ln(x+y)+x*cos(y)) is 1/(x+y)+cos(y)

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(\partial)/(\partial x)(ln(x+y)+x*cos(y))

\frac{\partial}{\partial x} (ln (x + y) + x \cdot cos (y))

\frac{1}{x + y} + cos (y)

Solution steps

\frac{\partial}{\partial x} (ln (x + y) + x cos (y))

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (ln (x + y)) + \frac{\partial}{\partial x} (x cos (y))

\frac{\partial}{\partial x} (ln (x + y)) = \frac{1}{x + y}

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

= \frac{1}{x + y} + cos (y)

\frac{d}{d x} (ln (x e^{- 9 x}))

\frac{\partial}{\partial x} (\frac{c - b}{a - x})

\frac{d}{d x} (m x^{n} - n x^{m})

y^{^{'}} = \frac{( 1 + x )}{x y ^{14}}

What is (\partial)/(\partial x)(ln(x+y)+x*cos(y)) ?
The answer to (\partial)/(\partial x)(ln(x+y)+x*cos(y)) is 1/(x+y)+cos(y)