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Implicit Differentiation: Finding Derivatives of Implicit Functions, Study notes of Analytical Geometry and Calculus

University of Washington (UW) - Seattle Analytical Geometry and Calculus

An explanation of implicit differentiation, a method used when a function cannot be expressed in terms of a single variable. It covers the steps to take the derivative of both sides of an equation with respect to a variable, solving for the derivative in terms of the variable and the implicit function, and evaluating the derivative at a point. Examples are given to illustrate the process.

Typology: Study notes

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5/2/07

Math 124 Lecture notes

Announcements & Reminders:

Work on Week 6 Hwk pbls. Bring questions to quiz section Thursday

Final exam: Saturday, June 2nd, 1:30-4:20 in KNE 210. More info: link on website.

Deadline for petition: May 11th.

Section 3.6: Implicit differentiation

Preparation:

1) Suppose 𝑦=𝑥3+ 5𝑥. Then 𝑦’= 3𝑥2+ 5.

Note: Here, by 𝑦’ I mean the derivative of y with respect to x, i.e. 𝑑𝑦

𝑑𝑥. The latter notation would

be better than the prime notation, because it makes it clearer which letter is the variable.

However, it is more cumbersome.

Consider the derivative of y^7.

 In terms of x, this is 𝑦7 ’= 𝑥3+ 5𝑥 7 ’= 7 𝑥3+ 5𝑥 6(3𝑥2+ 5)

 In terms of y, this is 𝑦7 ’= 7𝑦6𝑦′ (or, in the other notation: 𝑑

𝑑𝑥 𝑦7 = 7𝑦6𝑑𝑦

𝑑𝑥)

In both, we use Chain Rule.

2) Now suppose you don’t know the precise expression for y in terms of x, but you do know that y

depends on x via some relationship.

You do:

 𝑦2 ’=? Answer: 2yy’

(Can’t write it in terms of x because we don’t know what either y or y’ are in terms of x)

 𝑦3 ’=? Answer: 3𝑦2𝑦′

 1

𝑦 ’ =? Answer: −1

𝑦2𝑦′

 𝑒𝑦 ′=? Answer: 𝑒𝑦𝑦′

Note: in each of these, since we’re taking the derivative with respect to x not to y, we must

use Chain Rule, hence multiply by y’.

Here’s a tougher one:

 𝑥𝑒𝑦 ’=?

Since we have a product of two functions of x, namely x and ey, we must use the Product Rule:

𝑥𝑒𝑦 ’=𝑥′𝑒𝑦+𝑥 𝑒𝑦 ′= 1𝑒𝑦+𝑥𝑒𝑦𝑦′ =𝑒𝑦+𝑥𝑒𝑦𝑦′

Discover Study notes of Analytical Geometry and Calculus University of Washington (UW) - Seattle

Partial preview of the text

Download Implicit Differentiation: Finding Derivatives of Implicit Functions and more Study notes Analytical Geometry and Calculus in PDF only on Docsity!

Math 124 Lecture notes

Announcements & Reminders:

Work on Week 6 Hwk pbls. Bring questions to quiz section Thursday

Final exam: Saturday, June 2nd, 1:30-4:20 in KNE 210. More info: link on website. Deadline for petition: May 11th.

Section 3.6: Implicit differentiation

Preparation:

Suppose 𝑦 = 𝑥^3 + 5𝑥. Then 𝑦’ = 3𝑥^2 + 5. Note: Here, by 𝑦’ I mean the derivative of y with respect to x, i.e. 𝑑𝑦𝑑𝑥. The latter notation would be better than the prime notation, because it makes it clearer which letter is the variable. However, it is more cumbersome. Consider the derivative of y^7.  In terms of x, this is 𝑦^7 ’ = 𝑥^3 + 5𝑥 7 ’ = 7 𝑥^3 + 5𝑥 6 (3𝑥^2 + 5)  In terms of y, this is 𝑦^7 ’ = 7𝑦^6 𝑦′ (or, in the other notation: (^) 𝑑𝑥𝑑 𝑦^7 = 7𝑦^6 𝑑𝑦𝑑𝑥 )

In both, we use Chain Rule.

Now suppose you don’t know the precise expression for y in terms of x, but you do know that y depends on x via some relationship. You do:  𝑦^2 ’ =? Answer: 2yy’ (Can’t write it in terms of x because we don’t know what either y or y’ are in terms of x)  𝑦^3 ’ =? Answer: 3 𝑦^2 𝑦′  (^) 𝑦^1 ’ =? Answer: − 𝑦 12 𝑦′  𝑒𝑦^ ′^ =? Answer: 𝑒𝑦^ 𝑦′ Note: in each of these, since we’re taking the derivative with respect to x not to y, we must use Chain Rule, hence multiply by y’.

Here’s a tougher one:

 𝑥𝑒𝑦^ ’ =? Since we have a product of two functions of x, namely x and ey, we must use the Product Rule: 𝑥𝑒𝑦^ ’ = 𝑥′^ 𝑒𝑦^ + 𝑥 𝑒𝑦^ ′^ = 1𝑒𝑦^ + 𝑥𝑒𝑦^ 𝑦′ = 𝑒𝑦^ + 𝑥𝑒𝑦^ 𝑦′

When is all this useful?

Suppose you have a curve given by some equation in x and y and you cannot solve for y in terms of x (or is difficult to do so), but you’d still like to know the rate of change (derivative) of that curve at some point.

What to do?

METHOD of Implicit Differentiation:

We cannot write y as a function of x explicitly, but we can think of it as an implicit function of x, via the given equation.

Further, if two things are equal, then so are their derivatives. So:

STEP 1: Take the derivative of both sides of the equation with respect to x, treating any part containing y as practiced above (using Chain Rule).

You get an equation involving x, y and y’.

STEP 2: Solve the resulting equation for y’ in terms of x and y.

If y is actually a function of x and you can actually determine the formula for y in terms of x, replace it in the result so you’d get y’ in terms of x only. Otherwise, you have to be content with an expression depending both on the variable x and on the implicit function y.

STEP 3: If you need to evaluate the derivative at an actual point (a,b), then replace x=a, y=b into the formula you got for y’, then compute.

EXAMPLES:

Ex 1: Let’s compute the derivative y’ for the ellipse in the week 5 supplementary problem, namely:

𝒙𝟐 𝟒

+ 𝒚𝟐^ = 𝟏

Here, the variable is x and we consider y as an implicit function of x.

Step 1: Apply derivatives to both sides :

𝑥 2 4 +^ 𝑦

2 ′ = [ 1 ]’

Then evaluate them:

1 4 2 𝑥^ + 2𝑦𝑦′^ = 0

Step 2: Solve this equation for y’:

Implicit Differentiation: Finding Derivatives of Implicit Functions, Study notes of Analytical Geometry and Calculus

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Partial preview of the text

Download Implicit Differentiation: Finding Derivatives of Implicit Functions and more Study notes Analytical Geometry and Calculus in PDF only on Docsity!

+ 𝒚𝟐^ = 𝟏

2 ′ = [ 1 ]’