Question

How do you differentiate $\sin \left( arc\tan x \right)$?

Answer 1

Assume $f\left( x \right)=\sin x$ and $g\left( x \right)=\arctan x$ and write the given function as a composite function $y=f\left( g\left( x \right) \right)$. Write $\arctan x$ equal to $ta{{n}^{-1}}x$. Now, differentiate both sides of the function with respect to the variable x and use the chain rule of differentiation to find the derivative of $f\left( g\left( x \right) \right)$. Use the relation: - $\dfrac{d\left[ f\left( g\left( x \right) \right) \right]}{dx}=f'\left( g\left( x \right) \right)\times g'\left( x \right)$ to get the answer. Use the formula: - $\dfrac{d\left[ {{\tan }^{-1}}x \right]}{dx}=\left( \dfrac{1}{1+{{x}^{2}}} \right)$.

Complete step by step answer:
Here, we have been provided with the function $\sin \left( arc\tan x \right)$ and we are asked to find its derivative.
Now, let us assume this function as y, that means we have to find the value of $\dfrac{dy}{dx}$. So, we have,
$\because y=\sin \left( arc\tan x \right)$
We know that $arc\tan x={{\tan }^{-1}}x$, so we have,
$\Rightarrow y=\sin \left( {{\tan }^{-1}}x \right)$
We can convert the given function into a composite function because we have a combination of two functions. So, assuming $\sin x=f\left( x \right)$ and ${{\tan }^{-1}}x=g\left( x \right)$, we have,
$\Rightarrow y=f\left( g\left( x \right) \right)$
Differentiating both the sides with respect to x, we get,

& \Rightarrow \dfrac{dy}{dx}=\dfrac{d\left[ f\left( g\left( x \right) \right) \right]}{dx} \\\ & \Rightarrow \dfrac{dy}{dx}=f'\left( g\left( x \right) \right)\times g'\left( x \right) \\\ \end{aligned}$$ Here, we have applied the chain rule of differentiation in the R.H.S. and that is how the derivative of a composite function is determined. What we have to do in the next step is we have to find the derivative of $$f\left( g\left( x \right) \right)$$ with respect to $$g\left( x \right)$$, the derivative of $$g\left( x \right)$$ with respect to x and then take the product of these two derivatives obtained. So, mathematically, we have, $$\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left[ f\left( g\left( x \right) \right) \right]}{d\left[ g\left( x \right) \right]}\times \dfrac{d\left[ g\left( x \right) \right]}{dx}$$ Substituting the assumed functions, we have, $$\Rightarrow \dfrac{dy}{dx}=\dfrac{d\sin \left( {{\tan }^{-1}}x \right)}{d\left( {{\tan }^{-1}}x \right)}\times \dfrac{d\left( {{\tan }^{-1}}x \right)}{dx}$$ Using the formulas: - $$\dfrac{d\sin x}{dx}=\cos x$$ and $$\dfrac{d\left( {{\tan }^{-1}}x \right)}{dx}=\dfrac{1}{1+{{x}^{2}}}$$, we get, $$\begin{aligned} & \Rightarrow \dfrac{dy}{dx}=\cos \left( {{\tan }^{-1}}x \right)\times \dfrac{1}{1+{{x}^{2}}} \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{\cos \left( {{\tan }^{-1}}x \right)}{1+{{x}^{2}}} \\\ \end{aligned}$$ Hence, the derivative of the given function is $$\dfrac{\cos \left( {{\tan }^{-1}}x \right)}{1+{{x}^{2}}}$$. **Note:** One must remember the derivatives of basic functions like: - trigonometric functions, logarithmic functions, exponential functions etc to solve the above question. Here, in the above question we do not have any other method to solve the question. The given function must be converted into a composite function and then the chain rule must be applied. You must remember all the basic rules of differentiation like: - the product rule, chain rule, $$\dfrac{u}{v}$$ rule etc. as they are used everywhere in calculus.