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

The answer to (\partial)/(\partial x)(ln(e^{xy}cos(x))) is (ycos(x)-sin(x))/(cos(x))

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

\frac{1}{e ^{x y} cos ( x )} \frac{\partial}{\partial x} (e^{x y} cos (x))

= \frac{1}{e ^{x y} cos ( x )} \frac{\partial}{\partial x} (e^{x y} cos (x))

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

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

Simplify

\frac{1}{e ^{x y} cos ( x )} (y e^{y x} cos (x) - e^{y x} sin (x)) : \frac{y cos ( x ) - sin ( x )}{cos ( x )}

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

\frac{d}{d x} ((1 - x^{- 1})^{- 1})

y^{^{'}} = \frac{( 4 + 5 \cdot y )}{6}

\int x \cdot x^{\frac{1}{2}} d x

x \to 0 lim ((x^{2}) \cdot ln (x))

y^{^{'}} + \frac{1}{x} y = 0

What is (\partial)/(\partial x)(ln(e^{xy}cos(x))) ?
The answer to (\partial)/(\partial x)(ln(e^{xy}cos(x))) is (ycos(x)-sin(x))/(cos(x))