Find the derrivative of inverse trigonometric Function. y=1+x^(2)Arctan x-x

Recognize function components: Recognize the function components for differentiation. y = 1 + x^2 * arctan(x) - x We need to apply the product rule to x^2 * arctan(x) and the power rule to x.Apply derivative to each term: Apply the derivative to each term separately. Derivative of 1 is 0. Using the product rule for x^2 * arctan(x): (x^2)' * arctan(x) + x^2 * (arctan(x))' (x^2)' = 2x, and (arctan(x))' = 1/(1 + x^2) So, derivative of x^2 * arctan(x) = 2x * arctan(x) + x^2 * (1/(1 + x^2)) Derivative of -x is -1.Simplify the expression: Simplify the expression. dy/dx = 0 + 2x * arctan(x) + x^2/(1 + x^2) - 1Combine terms for derivative: Combine terms to finalize the derivative. dy/dx = 2x * arctan(x) + x^2/(1 + x^2) - 1

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Q. Find the derrivative of inverse trigonometric Function.

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y=1+x^{2} \operatorname{Arctan} x-x

Recognize function components: Recognize the function components for differentiation. $\newline$ $y = 1 + x^2 \cdot \arctan(x) - x$ $\newline$ We need to apply the product rule to $x^2 \cdot \arctan(x)$ and the power rule to $x$ .
Apply derivative to each term: Apply the derivative to each term separately. $\newline$ Derivative of $1$ is $0$ . $\newline$ Using the product rule for $x^2 \cdot \text{arctan}(x)$ : $(x^2)' \cdot \text{arctan}(x) + x^2 \cdot (\text{arctan}(x))'$ $\newline$ $(x^2)' = 2x$ , and $(\text{arctan}(x))' = \frac{1}{1 + x^2}$ $\newline$ So, derivative of $x^2 \cdot \text{arctan}(x)$ = $2x \cdot \text{arctan}(x) + x^2 \cdot \left(\frac{1}{1 + x^2}\right)$ $\newline$ Derivative of $-x$ is $-1$ .
Simplify the expression: Simplify the expression. $\frac{dy}{dx} = 0 + 2x \cdot \arctan(x) + \frac{x^2}{1 + x^2} - 1$
Combine terms for derivative: Combine terms to finalize the derivative. $\newline$ $\frac{dy}{dx} = 2x \cdot \arctan(x) + \frac{x^2}{1 + x^2} - 1$