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3.6 Derivatives as Rates of Change

Example

The position of an object moving horizontally on a straight line is

\[ s = f(t) = t^3 - 12t^2 + 45t \quad 0 \le t \le 7 \]

\( s > 0 \) corresponds to the right of the origin.

A horizontal number line with origin s=0 and an object labeled at a positive position s=f(t).
  • \( s > 0 \): right of origin
  • \( s < 0 \): left of origin
The position function tracks the displacement from the origin over time.

Graph of \( s = f(t) = t^3 - 12t^2 + 45t \)

A graph of position f(t) against time t, showing a cubic curve that rises, peaks at t=3, falls to t=5, then rises.

\( 0 < t < 3 \):

\( f(t) \) increasing, so getting farther away from origin. Moving RIGHT.

\( 3 < t < 5 \):

\( f(t) \) decreasing. Moving LEFT.

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The velocity is the rate of change of position.

\( ↳ \) derivative
\[ v(t) = s' = f'(t) = 3t^2 - 24t + 45 \]
A graph of velocity v(t) showing a parabola with roots at t=3 and t=5, positive outside and negative between them.

\( t \)-intercepts can be found by setting \( v(t) = 0 \).

Zero velocity ("stationary")
\[ 3t^2 - 24t + 45 = 0 \]\[ t^2 - 8t + 15 = 0 \]\[ (t - 3)(t - 5) = 0 \]\[ t = 3, \quad t = 5 \]

Motion Analysis from Velocity

  • \( 0 < t < 3 \): \( v(t) > 0 \) (above \( t \)-axis) means \( s = f(t) \) is increasing, so move RIGHT.
  • \( 3 < t < 5 \): \( v(t) < 0 \) means \( s = f(t) \) is decreasing, so move LEFT.
  • \( t > 5 \): move RIGHT again.

At \( t = 3, t = 5 \) object stops (\( v = 0 \)) when changing direction.

Velocity can be negative because it contains direction information, but speed cannot be negative.

speed = absolute value of velocity

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velocity is rate of change (derivative) of position

acceleration is the rate of change (derivative) of velocity

3.7 The Chain Rule (part 1)

\( y = (3x+5)^2 \quad y' = ? \)

one option: \( y = (3x+5)(3x+5) = 9x^2 + 30x + 25 \)

then \( y' = 18x + 30 \)

but notice we can't really do that with

\( y = (3x+5)^{17} \)

or \( y = (3x+5)^{3/4} \)

or \( y = (3x+5)^{-9} \)

we can use the Chain Rule

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basic idea: think of the function as a function of another function

then the derivative is the product of how each is affected

\( y = (3x+5)^2 \)

think of it as \( y = u^2 \) where \( u = 3x+5 \)

y as function of u

u as function of x

the derivative is product of how \( y \) is affected by \( u \) and how \( u \) is affected by \( x \)

\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]

how u affects y

how x affects u

another form: if \( y = f(g(x)) \)

then \( y' = f'(g(x)) \cdot g'(x) \)

\[ \frac{dy}{dx} = (2u) \cdot (3) = 6u = 6(3x+5) = 18x + 30 \]

same as before

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\[ y = (3x+5)^{17} \]
\[ y = u^{17} \quad \text{where } u = 3x+5 \]
\[ \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} = (17u^{16})(3) \]
\[ = 17(3x+5)^{16} \cdot (3) = 51(3x+5)^{16} \]

looks like the basic power rule if we pretend \( 3x+5 \) is just "\( x \)"

derivative of \( 3x+5 \) to account for the fact that \( 3x+5 \) is not just \( x \)

Another way to summarize:

\[ \frac{d}{dx} (\square)^n = n(\square)^{n-1} \cdot \frac{d}{dx}(\square) \]
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Example

\[ y = (3x^2+5)^{-3/4} \]
\[ y = u^{-3/4} \quad u = 3x^2+5 \]
\[ \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} = \left(-\frac{3}{4}u^{-7/4}\right)(6x) \]
\[ = -\frac{3}{4}(3x^2+5)^{-7/4}(6x) \]
\[ = -\frac{9}{2}x(3x^2+5)^{-7/4} \]

or:

\[ \frac{d}{dx} (3x^2+5)^{-3/4} \rightarrow \frac{d}{dx}(\square)^n = n(\square)^{n-1} \cdot \frac{d}{dx}(\square) \]

where \( \square \) is \( 3x^2+5 \)

\[ = -\frac{3}{4}(3x^2+5)^{-7/4} \cdot 6x \]
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Chain Rule Application

Example

\[ y = \sin^{50} x \]\[ y = (\sin x)^{50} \]

NOT the same as \( \sin(x^{50}) \)

Think of it as:

\[ y = (\square)^{50} \quad \implies \quad y' = 50(\square)^{49} \cdot \frac{d}{dx}(\square) \]

Applying this to the function:

\[ y' = 50(\sin x)^{49} \cdot \frac{d}{dx}(\sin x) \]\[ = 50(\sin x)^{49} \cdot \cos x \]
\[ 50 \cos x \sin^{49} x \]
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All the basic rules can be extended the same way:

\[ \frac{d}{dx} u^n = n u^{n-1} \cdot \frac{du}{dx} \]\[ \frac{d}{dx} \sin(u) = \cos(u) \cdot \frac{du}{dx} \]

From: \( \frac{d}{dx} \sin x = \cos x \)

Example: \( y = \sin(x^{50}) \)

\[ y' = \cos(x^{50}) \cdot \frac{d}{dx}(x^{50}) = \cos(x^{50}) \cdot 50x^{49} \]\[ = 50x^{49} \cos(x^{50}) \]

And so on.

Example: \( y = \sec(6x^2) \)

We know:

\[ \frac{d}{dx} \sec(x) = \sec(x) \tan(x) \]

So:

\[ \frac{d}{dx} \sec(u) = \sec(u) \tan(u) \cdot \frac{d}{dx}(u) \]
\[ y' = \sec(6x^2) \tan(6x^2) \cdot 12x \]
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Example

\[ y = e^{-2x^3} \]

We know:

\[ \frac{d}{dx}(e^x) = e^x \]

So:

\[ \frac{d}{dx} e^u = e^u \cdot \frac{du}{dx} \]

\[ y' = e^{-2x^3} \cdot \frac{d}{dx}(-2x^3) = e^{-2x^3} \cdot -6x^2 \]

\[ = -6x^2 e^{-2x^3} \]