Derivatives

The First Principles

The equation of a line is $y=mx+b$ , where $b$ is the y-intercept and $m$ is the slope. Calculating the slope is easy using any two distinct points on the line, $(x_1,y_1)$ and $(x_2,y_2)$

THEOREM

Slope

For $x_1\ne x_2$ , or equivalently $h\ne0$ ,

\begin{aligned} m&=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}\\[.65em] &=\frac{f(x_2)-f(x_1)}{x_2-x_1}=\frac{f(x+h)-f(x)}h \end{aligned}

Bring the second point toward the fixed point from either side. The green line is the tangent; the blue line joins the two points.

\displaystyle f(x)=x^2+1

\displaystyle T(t)=1.25+(1)(t-(0.5))

\displaystyle S(t)=1.25+(2)(t-(0.5))

Fixed input x0.5Horizontal change h1

\displaystyle \frac{f(0.5+(1))-f(0.5)}{1}=2

\displaystyle f'(0.5)=2(0.5)=1

What if we made the second point even closer to $(x,f(x))$ ?

As $h$ gets smaller in magnitude, the point $(x+h,f(x+h))$ approaches $(x,f(x))$ Where the derivative exists, the slope of the secant line approaches the slope of the tangent line. We can write this action mathematically as:

THEOREM

The First Principles

f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}h

If this limit exists as a finite number, its value equals the slope of the tangent line. This is the instantaneous rate of change of $f$ at $(x,f(x))$ This special limit is called the derivative.

If $y=f(x)$ , we can denote the derivative in a few different ways:

\frac{dy}{dx}\qquad y'\qquad f'(x)

When $f$ has a derivative at $(a,f(a))$ , we say it is differentiable at $x=a$ This requires the one-sided limits of the difference quotient to agree as finite numbers:

THEOREM

The First Principles: One-sided Slopes

\lim_{h\to0^-}\frac{f(a+h)-f(a)}h =\lim_{h\to0^+}\frac{f(a+h)-f(a)}h

QUESTION

For

\displaystyle f(x)=x^2

, what does

\displaystyle \displaystyle\lim_{h\to0}\dfrac{f(3+h)-f(3)}h

represent?

From the first principles, find the derivatives

(a) $\displaystyle f(x)=x^2$

(b) $\displaystyle f(x)=\sqrt{x+2}$

From The First Principles, find the derivative of $\displaystyle f(x) = {\dfrac{1}{x}}$ at $\displaystyle x = 5$

◀ Previous SectionLimits at Infinity Next Section ▶Differentiability and Continuity

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Derivatives

The First Principles

The equation of a line is $y=mx+b$ , where $b$ is the y-intercept and $m$ is the slope. Calculating the slope is easy using any two distinct points on the line, $(x_1,y_1)$ and $(x_2,y_2)$

THEOREM

Slope

For $x_1\ne x_2$ , or equivalently $h\ne0$ ,

\begin{aligned} m&=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}\\[.65em] &=\frac{f(x_2)-f(x_1)}{x_2-x_1}=\frac{f(x+h)-f(x)}h \end{aligned}

Bring the second point toward the fixed point from either side. The green line is the tangent; the blue line joins the two points.

\displaystyle f(x)=x^2+1

\displaystyle T(t)=1.25+(1)(t-(0.5))

\displaystyle S(t)=1.25+(2)(t-(0.5))

Fixed input x0.5Horizontal change h1

\displaystyle \frac{f(0.5+(1))-f(0.5)}{1}=2

\displaystyle f'(0.5)=2(0.5)=1

What if we made the second point even closer to $(x,f(x))$ ?

THEOREM

The First Principles

f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}h

If this limit exists as a finite number, its value equals the slope of the tangent line. This is the instantaneous rate of change of $f$ at $(x,f(x))$ This special limit is called the derivative.

If $y=f(x)$ , we can denote the derivative in a few different ways:

\frac{dy}{dx}\qquad y'\qquad f'(x)

When $f$ has a derivative at $(a,f(a))$ , we say it is differentiable at $x=a$ This requires the one-sided limits of the difference quotient to agree as finite numbers:

THEOREM

The First Principles: One-sided Slopes

\lim_{h\to0^-}\frac{f(a+h)-f(a)}h =\lim_{h\to0^+}\frac{f(a+h)-f(a)}h

QUESTION

For

\displaystyle f(x)=x^2

, what does

\displaystyle \displaystyle\lim_{h\to0}\dfrac{f(3+h)-f(3)}h

represent?

From the first principles, find the derivatives

(a) $\displaystyle f(x)=x^2$

(b) $\displaystyle f(x)=\sqrt{x+2}$

From The First Principles, find the derivative of $\displaystyle f(x) = {\dfrac{1}{x}}$ at $\displaystyle x = 5$

◀ Previous SectionLimits at Infinity Next Section ▶Differentiability and Continuity

The First Principles

The First Principles

Slope

From a secant to a tangent

The First Principles

The First Principles: One-sided Slopes

KNOWLEDGE QUIZ

Derive two functions from first principles

Find the reciprocal slope at five

The First Principles

The First Principles

Slope

From a secant to a tangent

The First Principles

The First Principles: One-sided Slopes

KNOWLEDGE QUIZ

Derive two functions from first principles

Find the reciprocal slope at five