Question

If it is given that$y=cos\left[ log\left( cot\;x \right) \right]$, then find the first order derivative$\dfrac{dy}{dx}$.

Answer 1

Hint: Apply the chain rule of differentiation that is given by$\dfrac{df\left( u \right)}{dx}=\dfrac{df}{du}\cdot \dfrac{du}{dx}$. Also, apply the formulas$\dfrac{d\left( \ln \left( u \right) \right)}{du}=\dfrac{1}{u},\dfrac{d\left( \cos \left( u \right) \right)}{du}=-\sin \left( u \right),\dfrac{d\left( \cot \left( x \right) \right)}{dx}=-{{\csc }^{2}}\left( x \right)$.

Complete step-by-step solution -
In the question, we have to find the first order derivative$\dfrac{dy}{dx}$ of the function$y=cos\left[ log\left( cot\;x \right) \right]$.
Now, we can write the derivative as;
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( \cos \left( \ln \left( \cot \left( x \right) \right) \right) \right)}{dx}$
Here, we have to apply the chain rule that is given as $\dfrac{df\left( u \right)}{dx}=\dfrac{df}{du}\cdot \dfrac{du}{dx}$
Here, $f=\cos \left( u \right),\;\;u=\ln \left( \cot \left( x \right) \right)$
So, the derivative will now is as follows:

& \Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( \cos \left( \ln \left( \cot \left( x \right) \right) \right) \right)}{dx} \\\ & \Rightarrow \Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( \cos \left( u \right) \right)}{du}\dfrac{d\left( \ln \left( \cot \left( x \right) \right) \right)}{dx} \\\ \end{aligned}$$ Next, we have to apply the formulas $$\dfrac{d\left( \ln \left( u \right) \right)}{du}=\dfrac{1}{u},\dfrac{d\left( \cos \left( u \right) \right)}{du}=-\sin \left( u \right),\dfrac{d\left( \cot \left( x \right) \right)}{dx}=-{{\csc }^{2}}\left( x \right)$$ and we will get: $$\begin{aligned} & \Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( \cos \left( u \right) \right)}{du}\dfrac{d\left( \ln \left( \cot \left( x \right) \right) \right)}{dx} \\\ & \Rightarrow \dfrac{dy}{dx}=\left( -\sin \left( u \right) \right)\dfrac{d\left( \ln \left( v \right) \right)}{dv}\dfrac{d\left( \cot \left( x \right) \right)}{dx}\because v=\cot \left( x \right),\dfrac{df\left( u \right)}{dx}=\dfrac{df}{dv}\cdot \dfrac{dv}{dx} \\\ & \Rightarrow \dfrac{dy}{dx}=\left( -\sin \left( u \right) \right)\left[ \dfrac{1}{v}\left( -{{\csc }^{2}}\left( x \right) \right) \right] \\\ & \Rightarrow \dfrac{dy}{dx}=\left( -\sin \left( u \right) \right)\left[ \dfrac{1}{\cot \left( x \right)}\left( -{{\csc }^{2}}\left( x \right) \right) \right]\because v=\cot \left( x \right) \\\ \end{aligned}$$ We will further simplify as follows: $$\begin{aligned} & \Rightarrow \dfrac{dy}{dx}=\left( -\sin \left( u \right) \right)\left[ \dfrac{\sin \left( x \right)}{-\cos \left( x \right){{\sin }^{2}}\left( x \right)} \right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\because \cot x=\dfrac{\cos \left( x \right)}{\sin \left( x \right)}\,\,\text{and}\,\,-{{\csc }^{2}}x=\dfrac{1}{-{{\sin }^{2}}\left( x \right)} \\\ & \Rightarrow \dfrac{dy}{dx}=\left( -\sin \left( u \right) \right)\left[ -\dfrac{1}{\cos \left( x \right)\sin \left( x \right)} \right] \\\ & \Rightarrow \dfrac{dy}{dx}=\left( -\sin \left( u \right) \right)\left[ -\dfrac{2}{2\cos \left( x \right)\sin \left( x \right)} \right] \\\ & \Rightarrow \dfrac{dy}{dx}=\left( -\sin \left( u \right) \right)\left[ -\dfrac{2}{\sin \left( 2x \right)} \right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\because 2\cos \left( x \right)\sin \left( x \right)=\sin \left( 2x \right) \\\ & \Rightarrow \dfrac{dy}{dx}=\left( -\sin \left( u \right) \right)\left[ -\dfrac{2}{\sin \left( 2x \right)} \right] \\\ & \Rightarrow \dfrac{dy}{dx}=\left( -\sin \left( u \right) \right)\left( -2\csc \left( 2x \right) \right) \\\ \end{aligned}$$ Next, substitute back$$u=\ln \left( \cot \left( x \right) \right)$$, we get the final expression of differentiation, as follows: $$\begin{aligned} & \Rightarrow \dfrac{dy}{dx}=\left( -\sin \left( u \right) \right)\left( -2\csc \left( 2x \right) \right) \\\ & \Rightarrow \dfrac{dy}{dx}=\left( -\sin \left( \ln \left( \cot \left( x \right) \right) \right) \right)\left( -2\csc \left( 2x \right) \right) \\\ & \Rightarrow \dfrac{dy}{dx}=2\sin \left( \ln \left( \cot \left( x \right) \right) \right)\csc \left( 2x \right) \\\ \end{aligned}$$ So the required derivative is $$\dfrac{dy}{dx}=2\sin \left( \ln \left( \cot \left( x \right) \right) \right)\csc \left( 2x \right)$$. Note: When finding the derivative of the log function, then it is important to bring the log function to the base if it is another base. Then apply the formula for the first order derivative of the log function which is given by $$\dfrac{d\left( \ln \left( u \right) \right)}{du}=\dfrac{1}{u}$$. In other words, we have to be careful if log function is given to the base e or to some other base. Hence, this formula is applicable if only we have the natural log function to the base e.