Find the derivative of $f(x) = x^2$.

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> [!TLDR]
> The derivative of $f(x) = x^2$ is found using the limit definition of the derivative. The key idea is to compute the slope of the tangent line at any point $x$, which gives $f'(x) = 2x$.

The derivative at a point measures the instantaneous rate of change — the slope of the tangent line. For a function like $f(x) = x^2$, which is a simple parabola, the slope changes at every point. To find a formula for that slope at any $x$, we use the limit definition:

$$f'(x) = \lim_{h 	o 0} \frac{f(x+h) - f(x)}{h}$$

This is the fundamental tool. Let’s apply it step by step.

1. **Set up the difference quotient.**  
   For $f(x) = x^2$, we have $f(x+h) = (x+h)^2$. So:
   $$\frac{f(x+h) - f(x)}{h} = \frac{(x+h)^2 - x^2}{h}$$

2. **Expand and simplify the numerator.**  
   $(x+h)^2 = x^2 + 2xh + h^2$. Subtract $x^2$:
   $$(x^2 + 2xh + h^2) - x^2 = 2xh + h^2$$
   So the quotient becomes:
   $$\frac{2xh + h^2}{h}$$

3. **Cancel the $h$ (provided $h 
eq 0$).**  
   Factor $h$ out of the numerator: $h(2x + h)$. Cancel with the denominator:
   $$\frac{h(2x + h)}{h} = 2x + h$$
   This is the slope of the secant line through $(x, f(x))$ and $(x+h, f(x+h))$.

4. **Take the limit as $h 	o 0$.**  
   As $h$ gets arbitrarily close to 0, the term $h$ vanishes, leaving:
   $$\lim_{h 	o 0} (2x + h) = 2x$$
   This limit is the derivative $f'(x)$. …

Q.Find the derivative of $f(x) = x^2$ .

Concept understanding — Rules of Differentiation

Q.Find the derivative of f(x)=x2f(x) = x^2f(x)=x2.

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Q.Find the derivative of $f(x) = x^2$ .

Unlock everything free for 14 days