Question

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

Answer 1

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,

& \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{aligned}$$ We can evaluate the $$\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.