# Dynamic Calculus

Home > The gradient function

## Differentiation from first principles (Part 2)

The applet below investigates the formal definition of the derivative and differentiation from first principles.

[You can reset the applet at any time by clicking the 'circular arrows' icon in the top right corner]

- Look closely at the definition of the function \(g(x)\). How does this relate to what you have learned about the formal definition of the derivative?
- Investigate the function of the 'slider' by moving it left and right. What do you observe about the position of \(y = g(x)\) as \(h\) approaches zero? What does this illustrate in relation to the gradient function of \(y = f(x)\)?
- In your book draw an accurate sketch of the function \(y = {x^2}\). Draw a tangent to the curve at the point where \(x = 2\). Calculate the gradient of the tangent.
- Find the gradient function for \(f(x) = {x^2}\). Check your solution by clicking the 'checkbox'. What is the gradient of the curve at the point where \(x = 2\)? Compare this answer with your answer from question (3).
- The gradient of the curve \(y = f(x)\) at the point \(P(c, f(c))\) is defined as the limiting value as \(x\) approaches \(c\) of \(\frac{{f(x) - f(c)}}{{x - c}}\), provided the limit exists. Thus \([f'(c)\) is the slope of the tangent to the curve \(y = f(x)\) at the point \(x = c\). By substituting \(x = c + h\) (where \(h\) may be positive or negative) in the definition, and noting that \(x\) approaches \(c\) as \(h\) approaches \(0\), verify that \(f'(c)\) is equal to the limit as \(h\) approaches \(0\) of \(\frac{{f(c + h) - f(c)}}{h}\).