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GRE Math: Study Guide & Test Prep27 chapters | 182 lessons | 16 flashcard sets

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Lesson Transcript

Instructor:
*Gerald Lemay*

Gerald has taught engineering, math and science and has a doctorate in electrical engineering.

Differentiation techniques include implicit differentiation. In this lesson we define the method of implicit differentiation and demonstrate this technique with numerous examples.

Let's say that our friend Gary is letting us know that he's going to be late for our 4:15 PM supper meeting, but all he says is that there's still lots of homework that he has to do. When asked to be clearer, Gary explicitly shares his schedule: at 4 PM he studies math, at 4:10 he takes a break, and at 4:20 he starts the homework. By explicitly stating his schedule, we know for sure Gary has no plans to be at the restaurant by 4:15.

In math differentiation problems, we often hear these same words: explicit and implicit. In this lesson we explain the implicit differentiation technique by comparing it to explicit differentiation. We then complete several examples using implicit differentiation. This might help Gary make it to the restaurant.

Given the equation *y*2 = *x* - 1, let's find the tangent line. We know taking the derivative of *y* with respect to *x* gives us the slope of the tangent line. If we first take the square root of both sides of this equation, we have:

This is the **explicit form** for the equation because *y* is isolated on the left-hand side and *x* appears only on the right-hand side.

Differentiating and solving for *y*', we get the equation below:

The first line says *y*' is the same as d*y* / d*x* is the same as D *y*. The prime notation, the d over d*x* notation, and the capital D notation are three ways to say the same thing: take the derivative with respect to *x*. It will be convenient to mix these notations.

The second line is the derivative of a square root, and the third line is the final answer where we are careful to avoid values for *x* undefined for *y*' (*x* = 0 leads to dividing by zero).

Differentiating an explicit form equation is called **explicit differentiation**.

Let's get the *y*' answer without first solving for *y*. Using the capital D notation for derivative, take the derivative of both sides of the equation:

D(*y* 2) = D(*x* -1) which gives:

2*y* *y* ' = D(x) - D(1) = 1 - 0, having used the chain rule for D(*y* 2). The derivative of a difference is the difference of the derivatives. Thus, D(*x* - 1) is the same as D(*x*) - D(1). The derivative of *x* is 1, and the derivative of a constant (the 1) is zero.

2*y* *y*' = 1, or *y*' = 1/(2 *y*). See how nicely the *y*' appears letting us easily solve for *y*'?

The **chain rule** is nicely expressed using the d d*x* notation:

See how the *y*' replaced the d*y* d*x*? Equivalent and more compact!

Substituting for *y* = square root(*x* - 1) we get:

The results are the same, but the second approach is very powerful when it's impossible or inconvenient to explicitly solve for *y*. This second approach is called **implicit differentiation**, which is essentially accomplished using the chain rule.

And one rule of effective study is to take breaks, including eating. Will Gary continue studying and miss supper?

Let's complicate the previous equation by mixing in more *x* and *y* terms:

(*x* - *y*)2 = *x* + 8*y* - 1. A plot of this curve looks like the image below with this equation:

Find the tangent line at the point *x* = 5, *y* = 1.25:

Differentiation gives the slope of the tangent line, but instead of trying to solve for *y* to do an explicit differentiation, we'll differentiate implicitly. First, take the derivative of both sides of the equation:

D (*x* - *y*)2 = D (*x* + 8*y* - 1)

Use the chain rule on the left-hand side, and on the right-hand side use the derivative of a sum is the sum of derivatives rule:

2(*x* - *y*)D(*x* - *y*) = D(*x*) + D(8*y*) - D(1)

Continue evaluating, and when we have a D(*y*), write this in prime notation as *y*':

2(*x* - *y*)(1 - *y*') = 1 + 8*y*' - 0

Grouping the *y*' terms on the right-hand side:

2(*x* - *y*) - 1 = *y*'(8 + 2(*x* - *y*))

Solving for *y*':

To complete this example, the equation of the tangent is *y* = m*x* + b where the slope m is *y*'. Substituting *x* = 5 and *y* = 1.25 into the *y*' result gives 0.42. Solving for b, we have b = *y* - m*x* = 1.25 - 0.42(5) = -0.85. Thus, the equation for the tangent to the curve at (5, 1.25) is *y* = 0.42*x* - 0.85. Plotting this line, we get the graph below:

A nice straight line and sometimes a straight line from the study hall to the restaurant is Gary's best path to get some food.

Let's do examples with exponentials and then with trig functions.

Take the derivative of both sides of the equation:

D e^(*x* + *y*) = D e^(e*x*) + D e^(2*y*)

Differentiate using the chain rule:

e^(*x* + *y*)D(*x* + *y*) = e^(3*x*)D(3*x*) + e^(2*y*)D(2*y*)

Evaluate the derivatives:

e^(*x* + *y*)(1 + *y*') = e^(3*x*)3 + e^(2*y*)2*y*'

Collect and factor the *y*' terms:

*y*'( e^(*x* + *y*) - 2e^(2*y*) ) = 3e^(3*x*) - e^(*x* + *y*)

Solve for *y*':

Differentiate both sides of the equation:

D cos *x* + D sin *y* = D sin(2*x* - 3*y*)

Evaluate the derivatives:

-sin *x* + cos (*y*)*y*' = cos(2*x* - 3*y*)D(2*x* - 3*y*) = cos(2*x* - 3*y*)(2 - 3*y*)

Collect and factor the *y*' terms:

*y*'(cos *y* + 3cos(2*x* - 3*y*) ) = 2cos(2*x* - 3*y*) + sin *x*

Solve for *y*':

Having all these math examples of implicit differentiation lets Gary feel confident to take an extended break and have some supper.

Let's take a few moments to review what we've learned in this lesson about the implicit differentiation technique. The **explicit form** of an equation has the *y* variable isolated on the left-hand side, and *x* appears only on the right-hand side. Differentiating an equation of this form is called **explicit differentiation**. For those cases where *y* isn't isolated, we can use **implicit differentiation**, which is essentially differentiating using the chain rule. The **chain rule** is nicely expressed using the d d*x* notation:

Now you should be able to use the implicit differentiation technique with ease!

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GRE Math: Study Guide & Test Prep27 chapters | 182 lessons | 16 flashcard sets

- Using Limits to Calculate the Derivative 8:11
- The Linear Properties of a Derivative 8:31
- Calculating Derivatives of Trigonometric Functions 7:20
- Linear Approximations Using Differentials: Definition & Examples 5:09
- Implicit Differentiation Technique, Formula & Examples 8:58
- Understanding Higher Order Derivatives Using Graphs 7:25
- Go to Calculating Derivatives

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