Calculus- Early Transcendentals, 2021a

150 Derivatives
Example 4.42: Equation and Derivative of Ellipse
Discuss the equation and derivative of the ellipse.
Solution. Consider all the points (x,y) that have the property that the distance from (x,y) to (x 1 ,y 1 ) plus
the distance from (x,y) to (x 2 ,y 2 ) is 2a (a is some constant). These points form an ellipse, which like a
circle is not a function but can be viewed as two functions pasted together. Since we know how to write
down the distance between two points, we can write down an implicit equation for the ellipse:
√
√
(x − x 1 ) 2 +(y − y 1 ) 2 + (x − x 2 ) 2 +(y − y 2 ) 2 = 2a.
Then we can use implicit differentiation to find the slope of the ellipse at any point, though the computation
is rather messy.
♣
Example 4.43: Derivative of Function defined Implicitly
Find dy
dx by implicit differentiation if 2x 3 + x 2 y − y 9 = 3x + 4.
Solution. Differentiating both sides with respect to x gives:
(
6x 2 + 2xy + x 2 dy )
− 9y 8 dy
dx dx = 3,
x 2 dy dy
− 9y8
dx dx = 3 − 6x2 − 2xy
(
x 2 − 9y 8) dy
dx = 3 − 6x2 − 2xy
dy
dx = 3 − 6x2 − 2xy
x 2 − 9y 8 .
♣
In the previous examples we had functions involving x and y, and we thought of y as a function of
x. In these problems we differentiated with respect to x. So when faced with x’s in the function we
differentiated as usual, but when faced with y’s we differentiated as usual except we multiplied by a dy
dx for
that term because we were using Chain Rule.
In the following example we will assume that both x and y are functions of t and want to differentiate
the equation with respect to t. This means that every time we differentiate an x we will be using the Chain
Rule, so we must multiply by dx
dy
dt
, and whenever we differentiate a y we multiply by
dt .