Let's try now to use implicit differentiation on our original equality to see if it works out: We must use the product rule again in the left side: Now we must substitute y as a function of x to compare it to our first result: And we got the same result, as expected. Return to Implicit Differentiation.

IMPLICIT DIFFERENTIATION PROBLEMS

In this section we will discuss implicit differentiation. Not every function can be explicitly written in terms of the independent variable, e.g. Y = f(x) and yet we will still need to know what f'(x) is. 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. To put simply because the calculators were designed as such. Lets have some history. First Casio calculators made before 1978 or so had no operator precedence. They calculated everything left to right so 1+2.3 gave 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

,

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 e^xy = e^4x - e^5y .

  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 x^2/3 + y^2/3 = 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 :

What is the difference between implicit and explicit differentiation?

1 Answer

It is a difference in how the function is presented before differentiating (or how the functions are presented).

Explanation:

#y = -3/5x+7/5# gives #y# explicitly as a function of #x#.

#3x+5y=7# gives exactly the same relationship between #x# and #y#, but the function is implicit (hidden) in the equation. To make the function explicit, we solve for #x#

6.1 Implicit Vs Explicitap Calculus Calculator

In #x^2+y^2=25#, #y# is not a function of #x#. However, there are two functions implicit in the equation. We can make the functions explicit by solving for #y#.

#y = +- sqrt(25-x^2)# is equivalent to the equation above and it has 2 functions that are not too difficult to make explicit:

#y=sqrt(25-x^2)# gives #y# as a function of #x# and

#y=-sqrt(25-x^2)# gives #y# as a different function of #x#.

We can differentiate either the implicit or explicit presentations.

6.1 Implicit Vs Explicitap Calculus Algebra

Differentiating implicitly (leaving the functions implicit) we get

#2x+2y dy/dx = 0##' '# so #' '##dy/dx = -x/y#

6.1 Implicit Vs Explicitap Calculus Solver

The #y# in the formula for the derivative is the price we pay for not making the function explicit. It replaces the explicit form of the function, whatever that may be.

For #y=sqrt(25-x^2)#, we get #dy/dx = - x/sqrt(25-x^2)# (use the power and chain rule), and

for #y= - sqrt(25-x^2)#, we get #dy/dx = x/sqrt(25-x^2)#.

The equation #y^5+4x^2y^2-3y+7x=28 # cannot be solved algebraically for #y#, (or anyway, some 5th degree equations cannot be solved) but there are several functions of #x# implicit in the equation. You can see them in the graph of the equation (shown below).

graph{y^5+4x^2y^2-3y+7x=28 [-7.14, 6.91, -4.66, 2.36]}

We can cut the graph into pieces, each of which is the graph of some function of #x# on some domain.

Implicit differentiation allow us to find the derivative(s) of #y# with respect to #x#without making the function(s) explicit. Doing that, we can find the slope of the line tangent to the graph at the point #(1,2)#.

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