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

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

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(\partial)/(\partial x)(ln(x+y)+xcos(y))

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

\int_{0}^{3} \frac{7 t}{( t - 4 ) ^{2}} d t

7 x^{e} + e^{9 x}

f (x) = \frac{x ^{2}}{5 + x}

\int (a^{2} - x^{2})^{\frac{5}{2}} d x

\int 4 x^{3} - 10 x d x

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