Question

How do you differentiate \[y=\sin x+{{x}^{2}}{{\tan }^{-1}}x\]?

Answer 1

Hint: To Solve the given question, we need to know some of the properties of differentiation, and derivatives of some functions. We should know the product rule of derivatives which states that \[\dfrac{d\left( f(x)g(x) \right)}{dx}=\dfrac{d\left( f(x) \right)}{dx}g(x)+f(x)\dfrac{d\left( g(x) \right)}{dx}\]. We should also know the derivatives of \[\sin x\], \[{{x}^{2}}\And {{\tan }^{-1}}x\]. The derivatives of these functions with respect to x are \[\cos x,2x\And \dfrac{1}{1+{{x}^{2}}}\] respectively. Also, we should know that derivatives can be separated over the addition of functions.

Complete step by step answer:
We are asked to differentiate the function \[y=\sin x+{{x}^{2}}{{\tan }^{-1}}x\]. We know that the derivative of a function can be separated over addition. Thus, we can evaluate the derivative as,
\[\begin{align}
  & \dfrac{dy}{dx}=\dfrac{d\left( \sin x+{{x}^{2}}{{\tan }^{-1}}x \right)}{dx} \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( \sin x \right)}{dx}+\dfrac{d\left( {{x}^{2}}{{\tan }^{-1}}x \right)}{dx} \\
\end{align}\]
We can evaluate the \[\dfrac{d\left( {{x}^{2}}{{\tan }^{-1}}x \right)}{dx}\] by using the product rule. The product rule of differentiation states that, \[\dfrac{d\left( f(x)g(x) \right)}{dx}=\dfrac{d\left( f(x) \right)}{dx}g(x)+f(x)\dfrac{d\left( g(x) \right)}{dx}\]. For this question, we have \[f(x)={{x}^{2}}\And g(x)={{\tan }^{-1}}x\]. Thus, using the product rule, we get \[\dfrac{d\left( {{x}^{2}}{{\tan }^{-1}}x \right)}{dx}=\dfrac{d\left( {{x}^{2}} \right)}{dx}{{\tan }^{-1}}x+{{x}^{2}}\dfrac{d\left( {{\tan }^{-1}}x \right)}{dx}\].
Thus, we can evaluate the derivate of given expression as follows,
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( \sin x \right)}{dx}+\dfrac{d\left( {{x}^{2}}{{\tan }^{-1}}x \right)}{dx}\]
Using the above product rule expression, we get
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( \sin x \right)}{dx}+\dfrac{d\left( {{x}^{2}} \right)}{dx}{{\tan }^{-1}}x+{{x}^{2}}\dfrac{d\left( {{\tan }^{-1}}x \right)}{dx}\]
We know that the derivatives of \[\sin x\], \[{{x}^{2}}\And {{\tan }^{-1}}x\] with respect to x are \[\cos x,2x\And \dfrac{1}{1+{{x}^{2}}}\] respectively. Substituting these expressions for the above derivatives, we get
\[\Rightarrow \dfrac{dy}{dx}=\cos x+2x{{\tan }^{-1}}x+{{x}^{2}}\dfrac{1}{1+{{x}^{2}}}\]
Simplifying the above expression, it can be written as
\[\Rightarrow \dfrac{dy}{dx}=\cos x+2x{{\tan }^{-1}}x+\dfrac{{{x}^{2}}}{1+{{x}^{2}}}\]
Thus, the derivative of the given expression is \[\cos x+2x{{\tan }^{-1}}x+\dfrac{{{x}^{2}}}{1+{{x}^{2}}}\].

Note:
To Solve these types of questions, we should know the different properties/ rules of differentiation like product rule, quotient rule, etc. Also, we should know the derivatives of different functions. For this question, we used the derivatives of function \[\sin x\], \[{{x}^{2}}\And {{\tan }^{-1}}x\].
We should also know the different methods like substitution to evaluate derivatives of complex function.