Differentiate Both Sides of the Equation
Once you have gotten the hang of implicit differentiation, it should not be difficult to take the derivative of both sides with respect to the variable t. This enables us to see how x and y vary with respect to time t. The only difference is that (x) = ,(y) = , and so on. Only (t) = 1 can be simplified, but this generally never occurs.
Example 1
Differentiate y^{2} + cos(x) = 4x^{2}y with respect to t.
Solution 1
Start with the equation.
y^{2} + cos(x) = 4x^{2}y
Differentiate both sides with respect to t.
(y^{2} + cos(x)) = (4x^{2}y)
Use the Chain Rule everywhere.
2y · (y) – sin(x) · (x) = 8x · (x) · y + (y) · 4x^{2}
Use (x) = and (y) = .
2y · – sin(x) · = 8xy · + 4x^{2}
Example 2
Differentiate e^{x} + y = y^{3} + √x with respect to t.
Solution 2
Start with the equation.
e^{x} + y = y^{3} + √x
Differentiate both sides with respect to t.
(e^{x} + y) = (y^{3} + √x)
Use the Chain Rule everywhere.
e^{x} · (x) + (y) = 3y^{2} · (y) + · (x)
Use (x) = and (y) = .
e^{x} · + +3y^{2} · + ·
The variables need not be x and y.
Example 3
Differentiate 3A + 4B^{2} = with respect to t.
Solution 3
(3A + 4B^{2}) =
Example 4
Differentiate A = π r^{2} with respect to t.
Solution 4
(A) = (π r^{2})
Don't forget that π is a constant, not a variable!
Just as is a rate, so are , , and so on. Because t usually represents time, = represents how fast y is changing over time. Thus, if A is a variable that represents an area, represents how fast that area. is increasing or decreasing.
Differentiating an equation with respect to t results in a new equation, which shows how the rates of change of the variables are related. For example, the area and radius of a circle are related by:
A = π r^{2}
If we differentiate with respect to t, we get:
= 2π r ·
If a circle is growing in size, this equation details how the rate at which the radius is changing, , relates to the rate at which the area is growing, .
Example 5
A rock thrown into a pond makes a circular ripple that travels at 4 feet per second. How fast is the area of the circle increasing when the circle has a radius of 12 feet?
Solution 5
We know that for circles, = 2π r · .And we know that the radius is increasing at the rate of = 4 feet per second, so when the radius is r = 12 feet, the area is increasing at:
= = 96π ≈ 301.6 square feet per second
Example 6
A spherical balloon is inflated with 40 cubic inches of air every second. When the radius is 12 inches, how fast is the radius of the balloon increasing? (Hint: The volume of a sphere with radius r is )
Solution 6
We know that the volume of the balloon is increasing at the rate of . We want to know what is When r = 12 inches. If we differentiate with respect to t, we get:
when we plug in and r = 12 in, we get:
The radius of the balloon is increasing at the very slow rate of ≈ 0.022 inches per second.
Example 7
Suppose the base of a triangle is increasing at a rate of 8 feet per minute while the height is decreasing by 1 foot every minute. How fast is the triangle's area changing when the height is 5 feet and the base is 20 feet?

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