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Calc Notes

The lecture focuses on understanding the derivative as a function, including its definition and graphical interpretation. It covers examples of computing derivatives, specifically for the functions f(x) = x^2 and f(x) = 1/x, illustrating how the derivative indicates the slope of the tangent line. Key takeaways emphasize that the derivative is a function itself and provides insights into the behavior of the original function regarding increasing and decreasing trends.

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9 views3 pages

Calc Notes

The lecture focuses on understanding the derivative as a function, including its definition and graphical interpretation. It covers examples of computing derivatives, specifically for the functions f(x) = x^2 and f(x) = 1/x, illustrating how the derivative indicates the slope of the tangent line. Key takeaways emphasize that the derivative is a function itself and provides insights into the behavior of the original function regarding increasing and decreasing trends.

Uploaded by

crazyamountsofspam
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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MATH 151 – Calculus I

Lecture 8: The Derivative as a Function


Date: September 19, 2024

Goals for Today:

• Understand the derivative as a function

• Practice computing derivatives from definition

• Graphical interpretation of the derivative

Quick Review:

• Definition of derivative at a point:


f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}
{h}f′(a)=limh→0hf(a+h)−f(a)

• This gives us the instantaneous rate of change of fff at aaa

Now: Derivative as a Function

We now look at:

f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}f′


(x)=h→0limhf(x+h)−f(x)

This defines a new function f′(x)f'(x)f′(x) that tells us the slope of the tangent line at every point
xxx in the domain of fff.

Example 1:

Let f(x)=x2f(x) = x^2f(x)=x2


Then:

f′
(x)=lim⁡h→0(x+h)2−x2h=lim⁡h→0x2+2xh+h2−x2h=lim⁡h→02xh+h2h=lim⁡h→0
(2x+h)=2xf'(x) = \lim_{h \to 0} \frac{(x + h)^2 - x^2}{h} = \lim_{h \to 0} \
frac{x^2 + 2xh + h^2 - x^2}{h} = \lim_{h \to 0} \frac{2xh + h^2}{h} = \
lim_{h \to 0} (2x + h) = 2xf′(x)=h→0limh(x+h)2−x2=h→0lim
hx2+2xh+h2−x2=h→0limh2xh+h2=h→0lim(2x+h)=2x

So f′(x)=2xf'(x) = 2xf′(x)=2x

Graphical Interpretation:

• f′(x)f'(x)f′(x) gives us the slope of the tangent line to f(x)f(x)f(x) at each


point

• If f(x)f(x)f(x) is increasing, f′(x)>0f'(x) > 0f′(x)>0

• If f(x)f(x)f(x) is decreasing, f′(x)<0f'(x) < 0f′(x)<0

• Critical points where f′(x)=0f'(x) = 0f′(x)=0 may be local max/min

Example 2:

Let f(x)=1xf(x) = \frac{1}{x}f(x)=x1


Using the limit definition:

f′(x)=lim⁡h→01x+h−1xh=lim⁡h→0x−(x+h)h(x)
(x+h)=lim⁡h→0−hhx(x+h)=−1x2f'(x) = \lim_{h \to 0} \frac{\frac{1}{x + h} -
\frac{1}{x}}{h} = \lim_{h \to 0} \frac{x - (x + h)}{h(x)(x + h)} = \lim_{h \
to 0} \frac{-h}{h x (x + h)} = \frac{-1}{x^2}f′(x)=h→0limhx+h1−x1
=h→0limh(x)(x+h)x−(x+h)=h→0limhx(x+h)−h=x2−1

So f′(x)=−1x2f'(x) = -\frac{1}{x^2}f′(x)=−x21

Conceptual Takeaways:

• The derivative is itself a function.

• We can graph f′(x)f'(x)f′(x) alongside f(x)f(x)f(x) to analyze behavior.

• Slope tells us about increasing/decreasing trends and concavity.

Homework:
• Complete Problems 3–15 odd from Section 2.3

• Quiz Friday: Derivatives from first principles

• Optional: Watch Prof. Gross’s video on tangent line intuition

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