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Question: What is the derivative of \(\arctan \left( \dfrac{y}{x} \right)\)?...

What is the derivative of arctan(yx)\arctan \left( \dfrac{y}{x} \right)?

Explanation

Solution

For solving this type of questions you should know about the differentiation of functions or the differentiation of inverse trigonometric functions. Because the value of the differentiation of trigonometric functions is also a trigonometric value but the differentiation of the inverse trigonometric functions are not trigonometric values. So, these are different from the differentiation of trigonometric functions in such a way.

Complete step by step solution:
In this question, we will obtain the derivative of tan1(yx)ta{{n}^{-1}}\left( \dfrac{y}{x} \right) . So, we have,
f(x)=tan1(yx)f\left( x \right)=ta{{n}^{-1}}\left( \dfrac{y}{x} \right)
Since, we know that ddx(tan1x)=11+x2\dfrac{d}{dx}\left( {{\tan }^{-1}}x \right)=\dfrac{1}{1+{{x}^{2}}},we can write it as,
ddx[tan1(yx)]=11+(yx)2.(ddx.yx)\dfrac{d}{dx}\left[ ta{{n}^{-1}}\left( \dfrac{y}{x} \right) \right]=\dfrac{1}{1+{{\left( \dfrac{y}{x} \right)}^{2}}}.\left( \dfrac{d}{dx}.\dfrac{y}{x} \right)
Since y is constant with respect to x, we can say
ddx.yx=yd(1x)dx ddx.yx=yx2 \begin{aligned} & \dfrac{d}{dx}.\dfrac{y}{x}=y\dfrac{d\left( \dfrac{1}{x} \right)}{dx} \\\ & \Rightarrow \dfrac{d}{dx}.\dfrac{y}{x}=-\dfrac{y}{{{x}^{2}}} \\\ \end{aligned}
And therefore, we can write
ddx[tan1(yx)]=11+(y2x2).(yx2)\Rightarrow \dfrac{d}{dx}\left[ ta{{n}^{-1}}\left( \dfrac{y}{x} \right) \right]=\dfrac{1}{1+\left( \dfrac{{{y}^{2}}}{{{x}^{2}}} \right)}.\left( \dfrac{-y}{{{x}^{2}}} \right)
By solving this equation, we can write it as follows,
ddx[tan1(yx)]=x2x2+y2(yx2)\dfrac{d}{dx}\left[ ta{{n}^{-1}}\left( \dfrac{y}{x} \right) \right]=\dfrac{{{x}^{2}}}{{{x}^{2}}+{{y}^{2}}}\left( \dfrac{-y}{{{x}^{2}}} \right)
We can further write it as,
ddx[tan1(yx)]=yx2+y2\dfrac{d}{dx}\left[ ta{{n}^{-1}}\left( \dfrac{y}{x} \right) \right]=\dfrac{-y}{{{x}^{2}}+{{y}^{2}}}
So, the derivative of arctan(yx)\arctan \left( \dfrac{y}{x} \right) is yx2+y2\dfrac{-y}{{{x}^{2}}+{{y}^{2}}}.
We can also solve this question by the chain rule. In this method we have to make both the functions as [f(g(x))]\left[ f\left( g\left( x \right) \right) \right] and the derivative of this will be as ddx[f(g(x))]=f(g(x))g(x)\dfrac{d}{dx}\left[ f\left( g\left( x \right) \right) \right]=f'\left( g\left( x \right) \right)g'\left( x \right), where f(x)=arctan(x)f\left( x \right)=\arctan \left( x \right) and g(x)=yxg\left( x \right)=\dfrac{y}{x}. And then we have to simplify this after differentiating to the function.

Note: For solving this question you should apply the direct method because it is a clear and simple method and this gives high accuracy to be the right answer. But if you use the chain rule, then that is a complicated method and that makes a high probability of your answer being wrong. But if the question is given to do it by chain rule, then you have to do it using that method only, otherwise you can use the direct method.