derivatives

$$‘‘‘latex\begin{array}{rl}\text{DEFINITIONS}& \text{The slope of the curve }y=f(x)\text{ at the point }P(x_0,f(x_0))\text{ is the}\\ & \text{number}\\ & m=\underset{h\to 0}{\lim }\frac{f(x_0+h)-f(x_0)}{h}\text{(provided the limit exists).}\\ & \text{The tangent line to the curve at }P\text{ is the line through }P\text{ with this slope.}\end{array}‘‘‘$$

Example

find the slope of the function’s graph at the given point. Then find an equation for the line tangent to the graph there.

• ƒ(x)= x2 + 1, (2, 5)

Step 1: Apply the Definition of Slope Using the limit definition: $m={\lim }_{h\to 0}\frac{f(x_0+h)-f(x_0)}{h}$ Step 2: Calculate f(x₀ + h) Given: $f(x)=x^2+1$

$f(x_0+h)=(x_0+h{)}^2+1$ $=x_0^2+2x_{0}h+h^2+1$

Step 3: Set Up the Difference Quotient

$\frac{f(x_0+h)-f(x_0)}{h}=\frac{(x_0^2+2x_{0}h+h^2+1)-(x_0^2+1)}{h}$

Step 4: Simplify the Numerator

$=\frac{x_0^2+2x_{0}h+h^2+1-x_0^2-1}{h}$ $=\frac{2x_{0}h+h^2}{h}$

$=\frac{h(2x_0+h)}{h}=2x_0+h$ (for h ≠ 0)

Step 5: Take the Limit

$m={\lim }_{h\to 0}(2x_0+h)=2x_0$

The slope at point (x₀, f(x₀)) is m = 2x₀

Step 7: Find the Tangent Line Equation

Using point-slope form: $y-y_1=m(x-x_1)$

At point $(x_0,x_0^2+1)$ with slope $m=2x_0$:

$y-(x_0^2+1)=2x_0(x-x_0)$ $y-x_0^2-1=2x_{0}x-2x_0^2$ $y=2x_{0}x-2x_0^2+x_0^2+1$ $y=2x_{0}x-x_0^2+1$

General tangent line equation: y = 2x₀x - x₀² + 1

At point (2, 5)

If we want the tangent line at x₀ = 2:

• Point: $(2,f(2))=(2,2^2+1)=(2,5)$
• Slope: $m=2(2)=4$
• Tangent line: $y=2(2)x-2^2+1=4x-4+1=4x-3$

At (2, 5): Slope = 4, Tangent line: y = 4x - 3

deeper

differentiation rules

derivatives of trigonometric functions