IMPLICIT DIFFERENTIATION PROBLEMS
- 6.1 Implicit Vs Explicitap Calculus 2nd Edition
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Instantaneous rate of change. Derivatives: chain rule and other advanced topics Implicit. Entry #6: 1.5 Limit Laws & The Squeeze Theorem Entry #7: 2.1 Differentiation Rules, Tangent Lines & Rates of Change Entry #8: 2.2 Product Rule, Quotient Rule & Trig Rules Entry #9: 2.3 Logarithmic Functions & Log Derivatives Entry #10: 2.4 The Chain Rule Entry #11: 3.1 Implicit Differentiation Entry #12: 3.2 Derivatives of Inverse Functions. Using a calculator, the value of to four decimal places is 3.0166. The value given by the linear approximation, 3.0167, is very close to the value obtained with a calculator, so it appears that using this linear approximation is a good way to estimate, at least for near 9.
The following problems require the use of implicit differentiation. Implicit differentiation is nothing more than a special case of the well-known chain rule for derivatives. The majority of differentiation problems in first-year calculus involve functions y written EXPLICITLY as functions of x . For example, if
What I want to show you in this video is that implicit differentiation will give you the same result as, I guess we can say, explicit differentiation when you can differentiate explicitly. So let's say that I have the relationship x times the square root of y is equal to 1.
,
then the derivative of y is
.
However, some functions y are written IMPLICITLY as functions of x . A familiar example of this is the equation
x2 + y2 = 25 ,
which represents a circle of radius five centered at the origin. Suppose that we wish to find the slope of the line tangent to the graph of this equation at the point (3, -4) .
How could we find the derivative of y in this instance ? One way is to first write y explicitly as a function of x . Thus,
x2 + y2 = 25 ,
y2 = 25 - x2 ,
and
,
where the positive square root represents the top semi-circle and the negative square root represents the bottom semi-circle. Since the point (3, -4) lies on the bottom semi-circle given by
,
the derivative of y is
,
i.e.,
.
Thus, the slope of the line tangent to the graph at the point (3, -4) is
.
Unfortunately, not every equation involving x and y can be solved explicitly for y . For the sake of illustration we will find the derivative of y WITHOUT writing y explicitly as a function of x . Recall that the derivative (D) of a function of x squared, (f(x))2 , can be found using the chain rule :
.
Since y symbolically represents a function of x, the derivative of y2 can be found in the same fashion :
.
Now begin with
x2 + y2 = 25 .
Differentiate both sides of the equation, getting
D ( x2 + y2 ) = D ( 25 ) ,
D ( x2 ) + D ( y2 ) = D ( 25 ) ,
and
2x + 2 y y' = 0 ,
so that
2 y y' = - 2x ,
and
,
i.e.,
.
Thus, the slope of the line tangent to the graph at the point (3, -4) is
.
This second method illustrates the process of implicit differentiation. It is important to note that the derivative expression for explicit differentiation involves x only, while the derivative expression for implicit differentiation may involve BOTH x AND y .
The following problems range in difficulty from average to challenging.
- PROBLEM 1 : Assume that y is a function of x . Find y' = dy/dx for x3 + y3 = 4 .
Click HERE to see a detailed solution to problem 1.
- PROBLEM 2 : Assume that y is a function of x . Find y' = dy/dx for (x-y)2 = x + y - 1 .
Click HERE to see a detailed solution to problem 2.
- PROBLEM 3 : Assume that y is a function of x . Find y' = dy/dx for .
Click HERE to see a detailed solution to problem 3.
- PROBLEM 4 : Assume that y is a function of x . Find y' = dy/dx for y = x2y3 + x3y2 .
Click HERE to see a detailed solution to problem 4.
- PROBLEM 5 : Assume that y is a function of x . Find y' = dy/dx for exy = e4x - e5y .
Click HERE to see a detailed solution to problem 5.
- PROBLEM 6 : Assume that y is a function of x . Find y' = dy/dx for .
Click HERE to see a detailed solution to problem 6.
- PROBLEM 7 : Assume that y is a function of x . Find y' = dy/dx for x=3 + sqrt{x^2+y^2} .
Click HERE to see a detailed solution to problem 7.
- PROBLEM 8 : Assume that y is a function of x . Find y' = dy/dx for .
Click HERE to see a detailed solution to problem 8.
- PROBLEM 9 : Assume that y is a function of x . Find y' = dy/dx for .
Click HERE to see a detailed solution to problem 9.
- PROBLEM 10 : Find an equation of the line tangent to the graph of (x2+y2)3 = 8x2y2 at the point (-1, 1) .
Click HERE to see a detailed solution to problem 10.
- PROBLEM 11 : Find an equation of the line tangent to the graph of x2 + (y-x)3 = 9 at x=1 .
Click HERE to see a detailed solution to problem 11.
- PROBLEM 12 : Find the slope and concavity of the graph of x2y + y4 = 4 + 2x at the point (-1, 1) .
Click HERE to see a detailed solution to problem 12.
- PROBLEM 13 : Consider the equation x2 + xy + y2 = 1 . Find equations for y' and y' in terms of x and y only.
Click HERE to see a detailed solution to problem 13.
- PROBLEM 14 : Find all points (x, y) on the graph of x2/3 + y2/3 = 8 (See diagram.) where lines tangent to the graph at (x, y) have slope -1 .
Click HERE to see a detailed solution to problem 14.
- PROBLEM 15 : The graph of x2 - xy + y2 = 3 is a 'tilted' ellipse (See diagram.). Among all points (x, y) on this graph, find the largest and smallest values of y . Among all points (x, y) on this graph, find the largest and smallest values of x .
Click HERE to see a detailed solution to problem 15.
- PROBLEM 16 : Find all points (x, y) on the graph of (x2+y2)2 = 2x2-2y2 (See diagram.) where y' = 0.
Click HERE to see a detailed solution to problem 16.
Click HERE to return to the original list of various types of calculus problems.
Your comments and suggestions are welcome. Please e-mail any correspondence to Duane Kouba by clicking on the following address :
Duane Kouba1998-06-23
Finding the derivative when you can't solve for y
You may like to read Introduction to Derivatives and Derivative Rules first.
Implicit vs Explicit
A function can be explicit or implicit:
Explicit: 'y = some function of x'. When we know x we can calculate y directly.
Implicit: 'some function of y and x equals something else'. Knowing x does not lead directly to y.
Example: A Circle
| Explicit Form | Implicit Form |
| y = ± √ (r2 − x2) | x2 + y2 = r2 |
| In this form, y is expressed as a function of x. | In this form, the function is expressed in terms of both y and x. |
The graph of x2 + y2 = 32
How to do Implicit Differentiation
- Differentiate with respect to x
- Collect all the dydx on one side
- Solve for dydx
6.1 Implicit Vs Explicitap Calculus 2nd Edition
Example: x2 + y2 = r2
Differentiate with respect to x:
ddx(x2) + ddx(y2) = ddx(r2)
Let's solve each term:
Which gives us:
2x + 2ydydx = 0
Collect all the dydx on one side
ydydx = −x
Solve for dydx:
dydx = −xy
The Chain Rule Using dydx
Let's look more closely at how ddx(y2) becomes 2ydydx
The Chain Rule says:
dudx = dudydydx
Substitute in u = y2:
ddx(y2) = ddy(y2)dydx
And then:
ddx(y2) = 2ydydx
Basically, all we did was differentiate with respect to y and multiply by dydx
Another common notation is to use ' to mean ddx
The Chain Rule Using '
The Chain Rule can also be written using ' notation:
f(g(x))' = f'(g(x))g'(x)
g(x) is our function 'y', so:
f(y)' = f'(y)y'
f(y) = y2, so f'(y) = 2y:
f(y)' = 2yy'
or alternatively: f(y)' = 2y dydx
Again, all we did was differentiate with respect to y and multiply by dydx
Explicit
Let's also find the derivative using the explicit form of the equation.
- To solve this explicitly, we can solve the equation for y
- Then differentiate
- Then substitute the equation for y again
6.1 Implicit Vs Explicitap Calculus Algebra
Example: x2 + y2 = r2
However, some functions y are written IMPLICITLY as functions of x . A familiar example of this is the equation
x2 + y2 = 25 ,
which represents a circle of radius five centered at the origin. Suppose that we wish to find the slope of the line tangent to the graph of this equation at the point (3, -4) .
How could we find the derivative of y in this instance ? One way is to first write y explicitly as a function of x . Thus,
x2 + y2 = 25 ,
y2 = 25 - x2 ,
and
,
where the positive square root represents the top semi-circle and the negative square root represents the bottom semi-circle. Since the point (3, -4) lies on the bottom semi-circle given by
,
the derivative of y is
,
i.e.,
.
Thus, the slope of the line tangent to the graph at the point (3, -4) is
.
Unfortunately, not every equation involving x and y can be solved explicitly for y . For the sake of illustration we will find the derivative of y WITHOUT writing y explicitly as a function of x . Recall that the derivative (D) of a function of x squared, (f(x))2 , can be found using the chain rule :
.
Since y symbolically represents a function of x, the derivative of y2 can be found in the same fashion :
.
Now begin with
x2 + y2 = 25 .
Differentiate both sides of the equation, getting
D ( x2 + y2 ) = D ( 25 ) ,
D ( x2 ) + D ( y2 ) = D ( 25 ) ,
and
2x + 2 y y' = 0 ,
so that
2 y y' = - 2x ,
and
,
i.e.,
.
Thus, the slope of the line tangent to the graph at the point (3, -4) is
.
This second method illustrates the process of implicit differentiation. It is important to note that the derivative expression for explicit differentiation involves x only, while the derivative expression for implicit differentiation may involve BOTH x AND y .
The following problems range in difficulty from average to challenging.
- PROBLEM 1 : Assume that y is a function of x . Find y' = dy/dx for x3 + y3 = 4 .
Click HERE to see a detailed solution to problem 1.
- PROBLEM 2 : Assume that y is a function of x . Find y' = dy/dx for (x-y)2 = x + y - 1 .
Click HERE to see a detailed solution to problem 2.
- PROBLEM 3 : Assume that y is a function of x . Find y' = dy/dx for .
Click HERE to see a detailed solution to problem 3.
- PROBLEM 4 : Assume that y is a function of x . Find y' = dy/dx for y = x2y3 + x3y2 .
Click HERE to see a detailed solution to problem 4.
- PROBLEM 5 : Assume that y is a function of x . Find y' = dy/dx for exy = e4x - e5y .
Click HERE to see a detailed solution to problem 5.
- PROBLEM 6 : Assume that y is a function of x . Find y' = dy/dx for .
Click HERE to see a detailed solution to problem 6.
- PROBLEM 7 : Assume that y is a function of x . Find y' = dy/dx for x=3 + sqrt{x^2+y^2} .
Click HERE to see a detailed solution to problem 7.
- PROBLEM 8 : Assume that y is a function of x . Find y' = dy/dx for .
Click HERE to see a detailed solution to problem 8.
- PROBLEM 9 : Assume that y is a function of x . Find y' = dy/dx for .
Click HERE to see a detailed solution to problem 9.
- PROBLEM 10 : Find an equation of the line tangent to the graph of (x2+y2)3 = 8x2y2 at the point (-1, 1) .
Click HERE to see a detailed solution to problem 10.
- PROBLEM 11 : Find an equation of the line tangent to the graph of x2 + (y-x)3 = 9 at x=1 .
Click HERE to see a detailed solution to problem 11.
- PROBLEM 12 : Find the slope and concavity of the graph of x2y + y4 = 4 + 2x at the point (-1, 1) .
Click HERE to see a detailed solution to problem 12.
- PROBLEM 13 : Consider the equation x2 + xy + y2 = 1 . Find equations for y' and y' in terms of x and y only.
Click HERE to see a detailed solution to problem 13.
- PROBLEM 14 : Find all points (x, y) on the graph of x2/3 + y2/3 = 8 (See diagram.) where lines tangent to the graph at (x, y) have slope -1 .
Click HERE to see a detailed solution to problem 14.
- PROBLEM 15 : The graph of x2 - xy + y2 = 3 is a 'tilted' ellipse (See diagram.). Among all points (x, y) on this graph, find the largest and smallest values of y . Among all points (x, y) on this graph, find the largest and smallest values of x .
Click HERE to see a detailed solution to problem 15.
- PROBLEM 16 : Find all points (x, y) on the graph of (x2+y2)2 = 2x2-2y2 (See diagram.) where y' = 0.
Click HERE to see a detailed solution to problem 16.
Click HERE to return to the original list of various types of calculus problems.
Your comments and suggestions are welcome. Please e-mail any correspondence to Duane Kouba by clicking on the following address :
Duane Kouba1998-06-23
Finding the derivative when you can't solve for y
You may like to read Introduction to Derivatives and Derivative Rules first.
Implicit vs Explicit
A function can be explicit or implicit:
Explicit: 'y = some function of x'. When we know x we can calculate y directly.
Implicit: 'some function of y and x equals something else'. Knowing x does not lead directly to y.
Example: A Circle
| Explicit Form | Implicit Form |
| y = ± √ (r2 − x2) | x2 + y2 = r2 |
| In this form, y is expressed as a function of x. | In this form, the function is expressed in terms of both y and x. |
The graph of x2 + y2 = 32
How to do Implicit Differentiation
- Differentiate with respect to x
- Collect all the dydx on one side
- Solve for dydx
6.1 Implicit Vs Explicitap Calculus 2nd Edition
Example: x2 + y2 = r2
Differentiate with respect to x:
ddx(x2) + ddx(y2) = ddx(r2)
Let's solve each term:
Which gives us:
2x + 2ydydx = 0
Collect all the dydx on one side
ydydx = −x
Solve for dydx:
dydx = −xy
The Chain Rule Using dydx
Let's look more closely at how ddx(y2) becomes 2ydydx
The Chain Rule says:
dudx = dudydydx
Substitute in u = y2:
ddx(y2) = ddy(y2)dydx
And then:
ddx(y2) = 2ydydx
Basically, all we did was differentiate with respect to y and multiply by dydx
Another common notation is to use ' to mean ddx
The Chain Rule Using '
The Chain Rule can also be written using ' notation:
f(g(x))' = f'(g(x))g'(x)
g(x) is our function 'y', so:
f(y)' = f'(y)y'
f(y) = y2, so f'(y) = 2y:
f(y)' = 2yy'
or alternatively: f(y)' = 2y dydx
Again, all we did was differentiate with respect to y and multiply by dydx
Explicit
Let's also find the derivative using the explicit form of the equation.
- To solve this explicitly, we can solve the equation for y
- Then differentiate
- Then substitute the equation for y again
6.1 Implicit Vs Explicitap Calculus Algebra
Example: x2 + y2 = r2
We get the same result this way!
You can try taking the derivative of the negative term yourself.
Chain Rule Again!
Yes, we used the Chain Rule again. Like this (note different letters, but same rule):
dydx = dydfdfdx
Substitute in f = (r2 − x2):
ddx(f½) = ddf(f½)ddx(r2 − x2)
Derivatives:
ddx(f½) = ½(f−½) (−2x)
And substitute back f = (r2 − x2):
ddx(r2 − x2)½ = ½((r2 − x2)−½) (−2x)
6.1 Implicit Vs Explicitap Calculus Calculator
And we simplified from there.
Using The Derivative
OK, so why find the derivative y' = −x/y ?
Well, for example, we can find the slope of a tangent line.
Example: what is the slope of a circle centered at the origin with a radius of 5 at the point (3,4)?
No problem, just substitute it into our equation:
dydx = −x/y
dydx = −3/4
And for bonus, the equation for the tangent line is:
y = −3/4 x + 25/4
Another Example
Sometimes the implicit way works where the explicit way is hard or impossible.
Example: 10x4 - 18xy2 + 10y3 = 48
How do we solve for y? We don't have to!
- First, differentiate with respect to x (use the Product Rule for the xy2 term).
- Then move all dy/dx terms to the left side.
- Solve for dy/dx
Like this:
(the middle term is explained below)
And we get:
| dydx = | 9y2 − 20x3 |
| 3(5y2 − 6xy) |
Product Rule
For the middle term we used the Product Rule: (fg)' = f g' + f' g
Because (y2)' = 2ydydx(we worked that out in a previous example)
Oh, and dxdx = 1, in other words x' = 1
Inverse Functions
Implicit differentiation can help us solve inverse functions.
6.1 Implicit Vs Explicitap Calculus Solver
The general pattern is:
- Start with the inverse equation in explicit form. Example: y = sin−1(x)
- Rewrite it in non-inverse mode: Example: x = sin(y)
- Differentiate this function with respect to x on both sides.
- Solve for dy/dx
As a final step we can try to simplify more by substituting the original equation.
An example will help:
Example: the inverse sine function y = sin−1(x)
We can also go one step further using the Pythagorean identity:
sin2 y + cos2 y = 1
cos y = √(1 − sin2 y )
And, because sin(y) = x (from above!), we get:
cos y = √(1 − x2)
Which leads to:
dydx= 1√(1 − x2)
Example: the derivative of square root √x
Note: this is the same answer we get using the Power Rule:
Summary
- To Implicitly derive a function (useful when a function can't easily be solved for y)
- Differentiate with respect to x
- Collect all the dy/dx on one side
- Solve for dy/dx
- To derive an inverse function, restate it without the inverse then use Implicit differentiation
