One of the most important applications of derivatives is understanding rates of change, particularly in motion problems involving position, velocity, and acceleration.
Given: \(s(t)\) = position at any given time
Average Velocity on \([a, b]\):
\[\text{Average velocity} = \frac{\text{total distance}}{\text{total time}} = \frac{s(b) - s(a)}{b - a}\]This is the slope of the secant line connecting points \((a, s(a))\) and \((b, s(b))\) on the position curve.
Instantaneous Velocity at \(a\):
\[v(a) = \lim_{b \to a} \frac{s(b) - s(a)}{b - a} = s'(a)\]This is the slope of the tangent line at \(x = a\) on the position curve.
Velocity Function:
\[v(t) = s'(t)\]Acceleration Function:
\[a(t) = v'(t) = s''(t)\]Graph Description: A coordinate plane showing a curved position function \(s(t)\) (blue curve). Two lines are drawn from a point at \(t = a\): a purple secant line connecting to a point at \(t = b\) (representing average velocity), and a blue tangent line at \(t = a\) (representing instantaneous velocity). Vertical dashed lines mark the positions at \(t = a\) and \(t = b\).
Given \(s(t) = t^3 - 9t^2 + 15t + 25\), find \(v(t)\) and \(a(t)\).
Solution:
Initial position: \(s(0) = 25\)
Velocity:
\[v(t) = s'(t) = \frac{d}{dt}(t^3 - 9t^2 + 15t + 25) = 3t^2 - 18t + 15\]Initial velocity: \(v(0) = 15\)
When velocity equals zero (object stops):
\[v(t) = 3t^2 - 18t + 15 = 3(t^2 - 6t + 5) = 3(t - 1)(t - 5) = 0\]Velocity is 0 at \(t = 1\) and \(t = 5\)
Acceleration:
\[a(t) = v'(t) = \frac{d}{dt}(3t^2 - 18t + 15) = 6t - 18\]Initial acceleration: \(a(0) = -18\)
When acceleration equals zero:
\[a(t) = 6t - 18 = 0 \implies t = 3\]Analysis:
Graph Description: Three stacked coordinate plane graphs showing the relationships between position, velocity, and acceleration over time. The top graph shows \(s(t)\) (blue parabolic curve) with a maximum around \(t = 1\) and minimum around \(t = 5\), with horizontal tangents marked at these points. The middle graph shows \(v(t)\) (purple parabola) crossing the x-axis at \(t = 1\) and \(t = 5\), with a minimum at \(t = 3\) where it has a horizontal tangent. The bottom graph shows \(a(t)\) (red line) crossing the x-axis at \(t = 3\). Annotations indicate where \(a(t) < 0\) (velocity decreasing) for \(t < 3\) and \(a(t) > 0\) (velocity increasing) for \(t > 3\).
So far, we've learned how to differentiate sums, differences, products, and quotients. But what about compositions of functions?
Find \(\frac{d}{dx}[\sin^2 x]\)
Using the Product Rule (inefficient):
\[\frac{d}{dx}[\sin^2 x] = \frac{d}{dx}[\sin x \cdot \sin x]\] \[= \frac{d}{dx}(\sin x) \cdot \sin x + \sin x \cdot \frac{d}{dx}(\sin x)\] \[= \cos x \cdot \sin x + \sin x \cdot \cos x = 2\sin x \cos x\]But what if we wanted \(\frac{d}{dx}[\sin^{75} x]\)? We would need to use the Product Rule 75 times!
Better approach: Think of \(\sin^2 x\) as \((\sin x)^2\), which is a composition of two functions:
We need a rule to handle this composition efficiently.
The Chain Rule:
If \(u = g(x)\) is the inside function and we want to find \(\frac{d}{dx}[f(g(x))]\), then:
\[\frac{d}{dx}[f(g(x))] = \frac{df}{du} \cdot \frac{du}{dx} = f'(g(x)) \cdot g'(x)\]In words: Take the derivative of the outside function (evaluated at the inside), then multiply by the derivative of the inside function.
Process:
Solution:
Inside function: \(u = \sin x\), so \(\frac{du}{dx} = \cos x\)
Outside function: \(u^{75}\), so \(\frac{d}{du}(u^{75}) = 75u^{74}\)
Using the Chain Rule:
\[\frac{d}{dx}[(\sin x)^{75}] = \frac{d}{du}(u^{75}) \cdot \frac{du}{dx}\] \[= 75u^{74} \cdot \cos x\]Replace \(u = \sin x\):
\[= 75(\sin x)^{74} \cdot \cos x\]Solution:
Inside function: \(u = x^{75}\), so \(\frac{du}{dx} = 75x^{74}\)
Outside function: \(\sin u\), so \(\frac{d}{du}(\sin u) = \cos u\)
Using the Chain Rule:
\[\frac{d}{dx}[\sin(x^{75})] = \cos u \cdot \frac{du}{dx}\]Replace \(u = x^{75}\):
\[= \cos(x^{75}) \cdot 75x^{74}\]Solution:
Inside function: \(u = 7x^3 + 2x - 9\), so \(\frac{du}{dx} = 21x^2 + 2\)
Outside function: \(u^{5/4}\), so \(\frac{d}{du}(u^{5/4}) = \frac{5}{4}u^{1/4}\)
Using the Chain Rule:
\[f'(x) = \frac{d}{dx}[u^{5/4}] = \frac{d}{du}(u^{5/4}) \cdot \frac{du}{dx}\] \[= \frac{5}{4}u^{1/4} \cdot (21x^2 + 2)\]Replace \(u = 7x^3 + 2x - 9\):
\[= \frac{5}{4}(7x^3 + 2x - 9)^{1/4} \cdot (21x^2 + 2)\]Solution:
Inside function: \(u = 8x^{11} - 23x + 1\), so \(\frac{du}{dx} = 88x^{10} - 23\)
Outside function: \(\tan u\), so \(\frac{d}{du}(\tan u) = \sec^2 u\)
Using the Chain Rule:
\[f'(x) = \frac{d}{dx}[\tan u] = \sec^2 u \cdot \frac{du}{dx}\]Replace \(u = 8x^{11} - 23x + 1\):
\[= \sec^2(8x^{11} - 23x + 1) \cdot (88x^{10} - 23)\]Solution:
This requires using the Chain Rule twice!
First application:
Inside function: \(u = \sin(x^{75})\)
Outside function: \(e^u\), so \(\frac{d}{du}(e^u) = e^u\)
\[\frac{d}{dx}[e^{\sin(x^{75})}] = e^u \cdot \frac{du}{dx}\]Second application (to find \(\frac{du}{dx}\)):
We need \(\frac{d}{dx}[\sin(x^{75})]\)
Inner function: \(x^{75}\), so \(\frac{d}{dx}(x^{75}) = 75x^{74}\)
Outer function: \(\sin u\), so \(\frac{d}{du}(\sin u) = \cos u\)
\[\frac{du}{dx} = \frac{d}{dx}[\sin(x^{75})] = \cos(x^{75}) \cdot 75x^{74}\]Putting it together:
\[\frac{d}{dx}[e^{\sin(x^{75})}] = e^{\sin(x^{75})} \cdot \cos(x^{75}) \cdot 75x^{74}\]Key Takeaway: The Chain Rule allows us to differentiate composite functions efficiently. The pattern is always: