Question

Find the derivative of ${{\tan }^{-1}}\left( \dfrac{\cos x}{1+\sin x} \right)$ with respect to ${{\sec }^{-1}}x$.

Answer 1

In this problem we need to find the derivative of the one function with respect to another function. So, we need to have the derivatives of the both the functions with respect to $x$. For this we will take the given first function as $v={{\tan }^{-1}}\left( \dfrac{\cos x}{1+\sin x} \right)$. To find the derivative of this function with respect to $x$ we will consider the part $\dfrac{\cos x}{1+\sin x}$, we will simplify this term by using some trigonometric functions after that we will derivate the simplified function. Now we will consider the second given function as $u={{\sec }^{-1}}x$. We will derivative this function with respect to $x$ and calculate the value of $\dfrac{du}{dx}$. After having the values of $\dfrac{dv}{dx}$, $\dfrac{du}{dx}$ we will calculate the required value which is $\dfrac{dv}{du}$ by dividing the value $\dfrac{dv}{dx}$ with $\dfrac{du}{dx}$.

Complete step-by-step solution:
Given functions are ${{\tan }^{-1}}\left( \dfrac{\cos x}{1+\sin x} \right)$, ${{\sec }^{-1}}x$.
Assuming the first given function as $v={{\tan }^{-1}}\left( \dfrac{\cos x}{1+\sin x} \right)$.
To find the derivative of the above function we are going to consider some part of the above function which is $\dfrac{\cos x}{1+\sin x}$. We are going to modify this part by using the trigonometric formulas $\sin \theta =\cos \left( \dfrac{\pi }{2}-\theta \right)$, $\cos \theta =\sin \left( \dfrac{\pi }{2}-\theta \right)$, then we will get
$\dfrac{\cos x}{1+\sin x}=\dfrac{\sin \left( \dfrac{\pi }{2}-x \right)}{1+\cos \left( \dfrac{\pi }{2}-x \right)}$
Applying the trigonometric formulas $\sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$, $1+\cos \theta =2{{\cos }^{2}}\dfrac{\theta }{2}$ in the above equation, then we will have
$\begin{aligned} & \dfrac{\cos x}{1+\sin x}=\dfrac{2\sin \left[ \dfrac{\left( \dfrac{\pi }{2}-x \right)}{2} \right]\cos \left[ \dfrac{\left( \dfrac{\pi }{2}-x \right)}{2} \right]}{2{{\cos }^{2}}\left[ \dfrac{\left( \dfrac{\pi }{2}-x \right)}{2} \right]} \\\ & \Rightarrow \dfrac{\cos x}{1+\sin x}=\dfrac{2\sin \left( \dfrac{\pi }{4}-\dfrac{x}{2} \right)\cos \left( \dfrac{\pi }{4}-\dfrac{x}{2} \right)}{2{{\cos }^{2}}\left( \dfrac{\pi }{4}-\dfrac{x}{2} \right)} \\\ \end{aligned}$
Cancelling the term $2\cos \left( \dfrac{\pi }{4}-\dfrac{x}{2} \right)$ which is in both numerator and denominator, then the above equation is modified as
$\dfrac{\cos x}{1+\sin x}=\dfrac{\sin \left( \dfrac{\pi }{4}-\dfrac{x}{2} \right)}{\cos \left( \dfrac{\pi }{4}-\dfrac{x}{2} \right)}$
We have the basic trigonometric definition $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$, from this we can write the above equation as
$\dfrac{\cos x}{1+\sin x}=\tan \left( \dfrac{\pi }{4}-\dfrac{x}{2} \right)$
Substituting the above in the assumed equation which is $v={{\tan }^{-1}}\left( \dfrac{\cos x}{1+\sin x} \right)$, then we will get
$v={{\tan }^{-1}}\left( \tan \left( \dfrac{\pi }{4}-\dfrac{x}{2} \right) \right)$
The functions ${{\tan }^{-1}}$, $\tan$ which are in the above equation will be get cancelled, then we will have
$v=\dfrac{\pi }{4}-\dfrac{x}{2}$
Now we have simplified form of the first given function which is ${{\tan }^{-1}}\left( \dfrac{\cos x}{1+\sin x} \right)$. Now we are going to find the derivative of this function with respect to $x$ by differentiating the simplified form, then we will get
$\dfrac{dv}{dx}=\dfrac{d}{dx}\left( \dfrac{\pi }{4}-\dfrac{x}{2} \right)$
Applying differentiation to each term individually and simplifying the equation by using the differentiation formula $\dfrac{d}{dx}\left( x \right)=1$, then we will have
$\begin{aligned} & \dfrac{dv}{dx}=\dfrac{d}{dx}\left( \dfrac{\pi }{4} \right)-\dfrac{1}{2}\times \dfrac{d}{dx}\left( x \right) \\\ & \Rightarrow \dfrac{dv}{dx}=0-\dfrac{1}{2}\times 1 \\\ & \Rightarrow \dfrac{dv}{dx}=-\dfrac{1}{2} \\\ \end{aligned}$
Now considering the given second function which is ${{\sec }^{-1}}x$.
Let us assume $u={{\sec }^{-1}}x$.
Differentiating the above equation with respect to $x$, then we will get
$\dfrac{du}{dx}=\dfrac{d}{dx}\left( {{\sec }^{-1}}x \right)$
We have the differentiation formula $\dfrac{d}{dx}\left( {{\sec }^{-1}}x \right)=\dfrac{1}{x\sqrt{{{x}^{2}}-1}}$, applying this formula in the above equation, then we will get
$\dfrac{du}{dx}=\dfrac{1}{x\sqrt{{{x}^{2}}-1}}$
Now we have the values of $\dfrac{dv}{dx}$, $\dfrac{du}{dx}$. But we need to find the derivative of first function$\left( v \right)$ with respect to second given function$\left( u \right)$ which is $\dfrac{dv}{du}$. We are going to calculate this value by dividing the value $\dfrac{dv}{dx}$ with $\dfrac{du}{dx}$, then we will get
$\begin{aligned} & \dfrac{dv}{du}=\dfrac{\dfrac{dv}{dx}}{\dfrac{du}{dx}} \\\ & \Rightarrow \dfrac{dv}{du}=\dfrac{-\dfrac{1}{2}}{\dfrac{1}{x\sqrt{{{x}^{2}}-1}}} \\\ & \Rightarrow \dfrac{dv}{du}=-\dfrac{x\sqrt{{{x}^{2}}-1}}{2} \\\ \end{aligned}$
Hence the derivative of the function ${{\tan }^{-1}}\left( \dfrac{\cos x}{1+\sin x} \right)$ with respect to ${{\sec }^{-1}}x$ is $-\dfrac{x\sqrt{{{x}^{2}}-1}}{2}$.

Note: In this problem we have simplified some part of the given first function which is $\dfrac{\cos x}{1+\sin x}$. We can also directly differentiate the first function with respect to $x$ and solve the problem. But when you decided to differentiate the first function ${{\tan }^{-1}}\left( \dfrac{\cos x}{1+\sin x} \right)$ with respect to $x$ directly we need to apply lot of differentiation formulas and it will become very hard to simplify the equation at some steps.