2.7. DERIVATIVES OF FUNCTIONS GIVEN IMPLICITLY
145
similar each time. After differentiating, we expand so that each side of the equation is a
sum of terms, some of which involve
dy
dx . Next, addition and subtraction are used to get
all terms involving
dy
dx on one side of the equation, with all remaining terms on the other.
Finally, we factor to get a single instance of
dy
dx , and then divide to solve for
dy
dx .
Note, too, that since
dy
dx is often a function of both x and y, we use the notation
dy
dx
(a,b)
to denote the evaluation of
dy
dx at the point (a, b). This is analogous to writing f ′ (a) when
f ′ depends on a single variable.
Finally, there is a big difference between writing
d
dx and
dy
dx . For example,
d
dx
[x
2 + y
2 ]
gives an instruction to take the derivative with respect to x of the quantity x 2 + y 2 ,
presumably where y is a function of x. On the other hand,
dy
dx
(x
2 + y
2 )
means the product of the derivative of y with respect to x with the quantity x 2 + y 2 .
Understanding this notational subtlety is essential.
The following activities present opportunities to explore several different problems
involving implicit differentiation.
Activity 2.19.
Consider the curve defined by the equation x = y 5 − 5y 3 + 4y, whose graph is pictured
in Figure 2.17.
(a) Explain why it is not possible to express y as an explicit function of x.
(b) Use implicit differentiation to find a formula for dy/dx.
(c) Use your result from part (b) to find an equation of the line tangent to the
graph of x = y 5 − 5y 3 + 4y at the point (0, 1).
(d) Use your result from part (b) to determine all of the points at which the graph
of x = y 5 − 5y 3 + 4y has a vertical tangent line.
⊳
Two natural questions to ask about any curve involve where the tangent line can be
vertical or horizontal. To be horizontal, the slope of the tangent line must be zero, while
to be vertical, the slope must be undefined. It is typically the case when differentiating
145
similar each time. After differentiating, we expand so that each side of the equation is a
sum of terms, some of which involve
dy
dx . Next, addition and subtraction are used to get
all terms involving
dy
dx on one side of the equation, with all remaining terms on the other.
Finally, we factor to get a single instance of
dy
dx , and then divide to solve for
dy
dx .
Note, too, that since
dy
dx is often a function of both x and y, we use the notation
dy
dx
(a,b)
to denote the evaluation of
dy
dx at the point (a, b). This is analogous to writing f ′ (a) when
f ′ depends on a single variable.
Finally, there is a big difference between writing
d
dx and
dy
dx . For example,
d
dx
[x
2 + y
2 ]
gives an instruction to take the derivative with respect to x of the quantity x 2 + y 2 ,
presumably where y is a function of x. On the other hand,
dy
dx
(x
2 + y
2 )
means the product of the derivative of y with respect to x with the quantity x 2 + y 2 .
Understanding this notational subtlety is essential.
The following activities present opportunities to explore several different problems
involving implicit differentiation.
Activity 2.19.
Consider the curve defined by the equation x = y 5 − 5y 3 + 4y, whose graph is pictured
in Figure 2.17.
(a) Explain why it is not possible to express y as an explicit function of x.
(b) Use implicit differentiation to find a formula for dy/dx.
(c) Use your result from part (b) to find an equation of the line tangent to the
graph of x = y 5 − 5y 3 + 4y at the point (0, 1).
(d) Use your result from part (b) to determine all of the points at which the graph
of x = y 5 − 5y 3 + 4y has a vertical tangent line.
⊳
Two natural questions to ask about any curve involve where the tangent line can be
vertical or horizontal. To be horizontal, the slope of the tangent line must be zero, while
to be vertical, the slope must be undefined. It is typically the case when differentiating