Tangent Slope on y = x²
Limits and Derivatives · Class XI
Slide a point x₀ along y = x². The tangent line uses slope 2x₀ , see it follow the curve.
- f(x₀)1
- f′(x₀)2
- Tangent y-intercept-1
- y = x²
- Tangent at x₀
Formulas in this lab
- f(x₀)
- f′(x₀)
- Tangent y-intercept
Frequently asked questions
▶Why is the derivative of $x^2$ equal to $2x$?
By first principles, $\frac{d}{dx}x^2 = \lim_{h\to 0}\frac{(x+h)^2 - x^2}{h} = \lim_{h\to 0}(2x + h) = 2x$. So the slope at $x = 3$ is $6$, and the slope at $x = -1$ is $-2$. It is the simplest power-rule case.
▶How do I use the derivative of $x^2$ lab?
Slide the point $x$ along the parabola $y = x^2$. The lab returns the slope $2x$ and the tangent equation $y = 2x_0(x - x_0) + x_0^2$. Try $x_0 = 2$: slope $4$ and tangent $y = 4x - 4$.
▶How does this preview the power rule?
The pattern generalizes: $\frac{d}{dx}x^n = nx^{n-1}$. So $\frac{d}{dx}x^3 = 3x^2$ and $\frac{d}{dx}x^{1/2} = \frac{1}{2}x^{-1/2}$. The $x^2$ case is the smallest non-trivial example, and once you trust it the formula slots in for all real $n$.
▶Where does the slope $2x$ appear in physics?
If $s(t) = t^2$ is the distance covered, then velocity is $v(t) = 2t$ — exactly the derivative. So a ball whose distance grows as $t^2$ has linearly-rising speed. The same pattern underlies free-fall kinematics: $s = \frac{1}{2}gt^2$, $v = gt$.