PYQ NEET- Continuity Differentiability & Derivatives

• 2019:

The derivative of $y = \tan^{-1}(\frac{1 - x}{1 + x})$ is $\frac{1}{1 + x^2}$.

To find the derivative, we first need to use the chain rule to differentiate the inside of the inverse tangent function. This gives us $\frac{d}{dx}(\frac{1 - x}{1 + x}) = \frac{-1}{(1 + x)^2}$. Then, we use the chain rule to differentiate the outside of the inverse tangent function. This gives us $\frac{d}{dx}(\tan^{-1}(\frac{1 - x}{1 + x})) = \frac{1}{1 + (\frac{1 - x}{1 + x})^2} \cdot \frac{-1}{(1 + x)^2}$.

• 2018:

The equation of the tangent to the curve $y = x^2 + 3x - 2$ at the point $(1, 2)$ is $y = -2x + 3$.

To find the equation of the tangent line, we first need to find the slope of the line. The slope of the tangent line is equal to the derivative of the curve at the