Find y' for the following equation. y = ln |sin x| .

Question:

Find {eq}y' {/eq} for the following equation.

{eq}y = \ln |\sin x| {/eq}.

Differentiation:

In mathematics, the differentiation is a process by which, the changing rate of something (like rate of change of velocity concerning time) can be obtained easily. For example, the acceleration of an object can be easily obtained by the differentiation of the velocity (in the variable form) of the object with respect to time.

Answer and Explanation:


Given function is shown below,

{eq}y=\ln |\sin x| {/eq}


Differentiate the above function with respect to x.

{eq}\dfrac{d}{dx}\left( y \right)=\dfrac{d}{dx}\left[ \ln |\sin x| \right]........................\left( 1 \right) {/eq}


The differentiation of the logarithmic function is shown below,

{eq}\text{Logrithmic}\ \text{function}\to \dfrac{d}{dx}\left[ \ln \left\{ u\left( x \right) \right\} \right]=\dfrac{1}{u(x)}\dfrac{d}{dx}\left[ u\left( x \right) \right] {/eq}

Apply, the differentiation of the logarithmic function in equation (1).

{eq}\begin{align*} \dfrac{dy}{dx} &=\dfrac{d}{dx}\left[ \ln |\sin x| \right] \\ \dfrac{dy}{dx} &=\dfrac{1}{|\sin x|}\dfrac{d}{dx}\left( |\sin x| \right)......................\left( 2 \right) \\ \end{align*} {/eq}


The differentiation of the modulus function is shown below,

{eq}\text{Modulus}\ \text{function}\to \dfrac{d}{dx}\left[ |u\left( x \right)| \right]=\dfrac{u\left( x \right)}{|u\left( x \right)|}\dfrac{d}{dx}\left[ u\left( x \right) \right] {/eq}

Apply, the differentiation of the modulus function in equation (2).

{eq}\begin{align*} \dfrac{dy}{dx} &=\dfrac{1}{|\sin x|}\dfrac{d}{dx}\left( |\sin x| \right) \\ \dfrac{dy}{dx} &=\dfrac{\dfrac{\sin x}{|\sin x|}\dfrac{d}{dx}\left( \sin x \right)}{|\sin x|}.........................\left( 3 \right) \\ \end{align*} {/eq}


The differentiation of trigonometric function is shown below,

{eq}\text{Trigonometric}\ \text{function}\to \dfrac{d}{dx}\left( \sin x \right)=\cos x {/eq}

Apply, the differentiation of the trigonometric function in equation (3).

{eq}\begin{align*} \dfrac{dy}{dx} &=\dfrac{\dfrac{\sin x}{|\sin x|}\dfrac{d}{dx}\left( \sin x \right)}{|\sin x|} \\ \dfrac{dy}{dx} &=\dfrac{\dfrac{\sin x}{|\sin x|}\left( \cos x \right)}{|\sin x|} \\ \dfrac{dy}{dx} &=\dfrac{\left( \sin x \right)\left( \cos x \right)}{\left( |\sin x| \right)\left( |\sin x| \right)}.......................\left( 4 \right) \\ \end{align*} {/eq}


The simple form of the modulus function is shown below,

{eq}\text{Modulus}\ \text{function}\to \text{ }\!\!|\!\!\text{ }\sin x|=\sqrt{{{\sin }^{2}}x} {/eq}

Apply simple form of the modulus function in equation (4).

{eq}\begin{align*} \dfrac{dy}{dx} &=\dfrac{\left( \sin x \right)\left( \cos x \right)}{\left( |\sin x| \right)\left( |\sin x| \right)} \\ \dfrac{dy}{dx} &=\dfrac{\left( \sin x \right)\left( \cos x \right)}{\left( \sqrt{{{\sin }^{2}}x} \right)\left( \sqrt{{{\sin }^{2}}x} \right)} \\ \dfrac{dy}{dx} &=\dfrac{\left( \sin x \right)\left( \cos x \right)}{\left( {{\sin }^{2}}x \right)} \\ \dfrac{dy}{dx} &=\dfrac{\cos x}{\sin x} \\ {y}' &=\cot x \\ \end{align*} {/eq}


Thus, the first derivative of the given function is {eq}{y}'=\cot x {/eq}.


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