What is the derivative of ln|e^{xy}cos(x|) ?

The derivative of ln|e^{xy}cos(x|) is (cos(x)(x\frac{dy)/(dx)+y)-sin(x)}{cos(x)}

What is the first derivative of ln|e^{xy}cos(x|) ?

The first derivative of ln|e^{xy}cos(x|) is (cos(x)(x\frac{dy)/(dx)+y)-sin(x)}{cos(x)}

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

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

Solution steps

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

Simplify

ln ∣ e^{x y} cos (x) ∣ : ln (e^{x y} ∣ cos (x) ∣)

= \frac{d}{d x} (ln (e^{x y} ∣ cos (x) ∣))

Apply the chain rule:

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

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

\frac{d}{d x} (e^{x y} ∣ cos (x) ∣) = e^{x y} ∣ cos (x) ∣ (y + x \frac{d y}{d x}) - \frac{e ^{x y} cos ( x ) sin ( x )}{∣ cos ( x ) ∣}

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

Simplify

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

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

2 y^{^{'}} - y = e^{\frac{x}{3}}

7 x + 2, at x = 2

5^{3 x}

\frac{d}{d x} ((x + 1)^{2} (x^{2} + 1)^{- 3})

\frac{\partial}{\partial x} (e^{x^{2} + y^{2}} \cdot 2 x)

What is the derivative of ln|e^{xy}cos(x|) ?
The derivative of ln|e^{xy}cos(x|) is (cos(x)(x\frac{dy)/(dx)+y)-sin(x)}{cos(x)}
What is the first derivative of ln|e^{xy}cos(x|) ?
The first derivative of ln|e^{xy}cos(x|) is (cos(x)(x\frac{dy)/(dx)+y)-sin(x)}{cos(x)}