qa16

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course Mth 271

11/8 830p

16. Implicit differentiation.

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Question: `q001. Suppose that y is a function of x. Then the derivative of y with respect to x is y '. The derivative of x itself, with respect to x, is by the power function rule 1, x being x^1. What therefore would be the derivative of the expression x^2 y?

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Your solution:

(first)(derivative of 2nd) + (second)(derivative of 1st)

x^2* 1 + y(2x)

x^2 + 2xy

confidence rating #$&*:2

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Given Solution:

`aBy the product rule we have (x^2 y) ' = (x^2) ' y + x^2 * y ' = 2 x y + x^2 y ' .

STUDENT QUESTION

I understand this concept except for where the 2xy came from?

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Self-critique Rating:ent: 1

INSTRUCTOR RESPONSE

(x^2 y) ' = (x^2) ' y + x^2 * y '

The 2 x y comes from (x^2) ' y:

(x^2) ' = 2x so (x^2)' y = 2 x y.

Thus (x^2) ' y + x^2 * y ' = 2 x y + x^2 y '

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Self-critique (if necessary):

If you split the functions into x^2 and y, I thought the derivative of y would be 1, but it looks like it is y. If the derivative of 3x is 3 then why isn’t the derivative of 1y = 1?

@&

Very good question.

The derivative of y with respect to y would indeed by 1, but the derivative here is taken with respect to x.

y is implicitly a function of x. So its derivative with respect to x is dy/dx, or y '.

On the other hand 3x is an explicit function of x, whose derivative with respect to x is 3.

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Self-critique Rating:2

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Question: `q002. Note that if y is a function of x, then y^3 is a composite function, evaluated from a given value of x by the chain of calculations that starts with x, then follows the rule for y to get a value, then cubes this value to get y^3. To find the derivative of y^3 we would then need to apply the chain rule. If we let y' stand for the derivative of y with respect to x, what would be the derivative of the expression y^3 with respect to x?

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Your solution:

y(x) = y(x)^3

y’(3y^2)

confidence rating #$&*:2

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Given Solution:

`aThe derivative of y^3 with respect to x can be expressed more explicitly by writing y^3 as y(x)^3. This reinforces the idea that y is a function of x and that we have a composite f(y(x)) of the function f(z) = z^3 with the function y(x). The chain rule tells us that the derivative of this function must be

(f ( y(x) ) )' = y ' (x) * f ' (y(x)),

in this case with f ' (z) = (z^3) ' = 3 z^2.

The chain rule thus gives us (y(x)^3) ' = y ' (x) * f ' (y(x) ) = y ' (x) * 3 * ( y(x) ) ^2.

In shorthand notation, (y^3) ' = y ' * 3 y^2.

This shows how the y ' comes about in implicit differentiation.

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Self-critique (if necessary):

My original answer was 2y’(6y^2) but I was using the product rule instead of answering the question you asked.

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Self-critique Rating:3

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Question: `q003. If y is a function of x, then what is the derivative of the expression x^2 y^3?

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Your solution:

first(derivative of 2nd) + (second)(derivative of 1st)

x^2(3y^2) + y^3(2x)= 3x^2y^2+ 2xy^3

confidence rating #$&*:2

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Given Solution:

`aThe derivative of x^2 y^3, with respect to x, is

(x^2 y^3) ' = (x^2)' * y^3 + x^2 * (y^3) ' = 2x * y^3 + x^2 * [ y ' * 3 y^2 ] = 2 x y^3 + 3 x^2 y^2 y '.

Note that when the expression is simplified, the y ' is conventionally placed at the end of any term of which is a factor.

STUDENT QUESTION

Ok I understand everything until the last part of the answer. How come the 3 only got combined with the x^2?

INSTRUCTOR RESPONSE

a * b * c = c * b * a = b * c * a = etc.. The order in which the quantities are listed in a string of multiplications doesn't matter.

So x^2 * [ y ' * 3 y^2 ] means the same thing as 3 y^2 * x^2 * y ' which means the same as 3 x^2 y^2 y '.

STUDENT QUESTION

I am confused about the last part of this problem why is g` not just 3y^2, I do not understand [ y ' * 3 y^2 ] why is this

multiplied by y`?

I see were the 3y^2 originated from but why is there a y`?

INSTRUCTOR RESPONSE

y is a function of x. So y^3 is a composite function, as shown in the preceding question.

Your answer to that question was that the derivative of (y(x))^3 was y`(x)*3*(y(x))^2.

Abbreviated, your result is y ' * 3 y^2, the same as 3 y^2 y '.

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Self-critique (if necessary):

so when we say that y is a function of x, that means we add the y’ to the end. I thought that it was implicit when dealing functions in the form of f(x) = x or y = x… that y is a function of x.

@&

y is indeed assumed to be an implicitly defined function of x.

The derivative of y with respect to x is dy/dx, or y '.

This makes y^3 a composite function of x. This might be clearer if we write y(x) instead of just y. Then instead of y^3 we would write y(x)^3. It should be clear that this is the composite of the cubing function and the function y(x).

This composite could be expressed as the composite of f(u) = u^3 and u = y(x).

f ' (u) = 3 u^2, so f ' (y(x)) = 3 * y(x)^2.

The derivative of this composite would therefore be y ' (x) * f ' (y(x) ) = y ' (x) * 3 y(x)^2. In abbreviated notation this is y ' * 3 y^2, written more grammatically as 3 y ' y^2.

*@

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Self-critique Rating:2

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Question: `q004. The equation 2x^2 y + 7 x = 9 can easily be solved for y. What is the result?

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Your solution:

y = (9-7x)/2x^2

confidence rating #$&*:2

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Given Solution:

`aStarting with 2 x^2 y + 7 x = 9 we first subtract seven x from both sides to obtain

2 x^2 y = 9 - 7 x. We then divide both sides by 2 x^2 to obtain

y = (9 - 7 x ) / (2 x^2), or if we prefer

y = 9 / (2 x^2 ) - 7 / ( 2 x ).

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Self-critique (if necessary):

n/a

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Self-critique Rating:3

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Question: `q005. Using the form y = 9 / (2 x^2 ) - 7 / ( 2 x ), what is y and what is y ' when x = 1?

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Your solution:

When x = 1, y=1

This actually solves easier to leave it in the first form: (9-7x)/(2x^2). ( I did make sure I came up with the right answer before making that comment!) It is easier to separate the parts into

[(bottom)/(derivative of top) - (top)(derivative of bottom)]/bottom^2

which gives

[(2x^2)(-7)- (9-7x)(4x)]/(2x^2)^2 which eventually simplifies to (7x-18)/-2x^2.

When you plug x=1 into this you get -11/2

confidence rating #$&*:3

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Given Solution:

`ay ' = 9/2 * ( 1/x^2) ' - 7/2 ( 1 / x) ' = 9/2 * (-2 / x^3) - 7/2 * (-1 / x^2) = -9 / x^3 + 7 / (2 x^2).

So when x = 1 we have

y = 9 / (2 * 1^2 ) - 7 / ( 2 * 1 ) = 9/2 - 7/2 = 1 and

y ' = -9 / 1^3 + 7 / (2 * 1^2) = -9 + 7/2 = -11 / 2.

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Self-critique (if necessary):n/a

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Self-critique Rating: 3

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Question: `q006. The preceding calculation could have been obtained without solving the original equation 2x^2 y + 7 x = 9 for y. We could have taken the derivative of both sides of the equation to get 2 ( (x^2) ' y + x^2 y ' ) + 7 = 0, or 2 ( 2x y + x^2 y ' ) + 7 = 0.

Complete the simplification of this equation, then solve for y ' .

Note that when x = 1 we have y = 1. Substitute x = 1, y = 1 into the equation for y ' and see what you get for y '.

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Your solution:

2xy + x^2y’ = -7/2

y’ = [(-7/2) - 2xy]/x^2

[(-7/2) - 2(1)(1)]/ 1^2 = -7/2

confidence rating #$&*:3

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Given Solution:

`aStarting with 2 ( 2x y + x^2 y ' ) + 7 = 0 we divide both sides by 2 to obtain

2x y + x^2 y ' + 7/2 = 0. Subtracting 2 x y + 7 / 2 from both sides we obtain

x^2 y' = - 2 x y - 7 / 2.

Dividing both sides by x^2 we end up with

y ' = (- 2 x y - 7 / 2) / x^2 = -2 y / x - 7 / ( 2 x^2).

Substituting x = 1, y = 1 we obtain

y ' = -2 * 1 / 1 - 7 / ( 2 * 1^2) = - 11 / 2.

Note that this agrees with the result y ' = -11/2 obtained when we solved the original equation and took its derivative. This shows that it is not always necessary to explicitly solve the original equation for y. This is a good thing because it is not always easy, in fact not always possible, to solve a given equation for y.

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Self-critique (if necessary):

n/a

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Self-critique Rating:3

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Question: `q007. Follow the procedure of the preceding problem to determine the value of y ' when x = 1 and y = 2, for the equation 2 x^2 y^3 - 3 x y^2 - 4 = 0. Also validate the fact that x = 1, y = 2 is a solution to the equation, because if this isn't a solution it makes no sense to ask a question about the equation for these values of x and y.

Solution:

Easy to validate that x= 1, y=2 is a solution:

2(1)^2*(2)^3 - 3(1)(2^2) - 4 = 0

Not so easy - as best I can determine

y’ = (2x^2)(3y^2) + (y^3)(4x) + (-3x)(2y) + y^2(-3) which simplifies to

6(1)^2(2^2)+ 4(1)(2^3) - 6(1)(2) - 2(2^2) = 32

@&

@&

You get (2x^2)(3 y^2) from the term (2x^2) (y^3) ' .

y^3 means y(x)^3, which is a composite function of x with derivative y ' (x) * 3 y(x)^2, which you would write as 3 y^2 y '.

So (2 x^2) (y^3)' = (2x^2) * (3 y^2 y ') = 6 x^2 y^2 y '.

*@

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@&

Similarly your term -3x * 2y is obtained from

(3x) ( y^2) '

which by similar reasoning can be written

(3x) * ( 2 y y ' ) = 6 x y y '.

*@

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Self-critique (if necessary):not sure this is right but don’t know what else to do with it

@&

Check my notes. You're close, but you have to treat y as a function of x, which will give you a factor of y ' on some of your terms.

*@

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Self-critique Rating:2

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Question: `qNote that this time we do not need to solve the equation explicitly for y. This is a good thing because this equation is cubic in y, and while there is a formula (rather a set of formulas) to do this it is a lengthy and messy process.

The derivative of the equation 2 x^2 y^3 - 3 x y^2 - 4 = 0 is (2 x^2 y^3) ' - (3 x y^2) ' - (4)' = 0. Noting that 4 is a constant, we see that (4)' = 0. The derivative of the equation therefore becomes

(2 x^2) ' * y^3 + 2 x^2 * ( y^3) ' - (3 x ) ' * y^2 - 3 x * (y ^2 ) ' = 0, or

4 x y^3 + 6 x^2 y^2 y' - 3 y^2 - 6xy y' = 0. Subtracting from both sides the terms which do not contain y ' we get

6 x^2 y^2 y ' - 6 x y y ' = - 4 x * y^3 + 3 y^2. Factoring out the y ' on the left-hand side we have

y ' ( 6 x^2 y^2 - 6 x y) = -4 x y^3 + 3 y^2. Dividing both sides by the coefficient of y ':

y ' = (- 4 x * y^3 + 3 y^2) / ( 6 x^2 y^2 - 6 x y ) . The numerator and denominator have common factor y so we end up with

y' = (- 4 x * y^2 + 3 y) / ( 6 x^2 y - 6 x ).

Now we see that if x = 1 and y = 2 the equation 2 x^2 y^3 - 3 x y^2 - 4 = 0 gives us

2 * 1^2 * 2^3 - 3 * 1 * 2^2 - 4 = 0, or

16 - 12 - 4 = 0, which is true.

Substituting x = 1 and y = 2 into the equation y' = (- 4 x * y^2 + 3 y) / ( 6 x^2 y - 6 x ) we get

y ' = (- 4 * 1 * 2^2 + 3 * 2) / ( 6 * 1^2 * 2 - 6 * 1 ) =

(-16 + 6) / (12 - 6) = -10 / 6 = -5/3 or -1.66... .

STUDENT QUESTION

I see what was done in the solution once I read that I did not have to solve for y. My approach to this was to solve for y

and that did not turn out well. My question is how do I know when to solve for y and when not to?

INSTRUCTOR RESPONSE

Once you're used to implicit differentiation, unless otherwise stated, then you can make that decision for yourself.

Of course for most expressions you won't be able to solve for y, so the decision is already made by the problem.

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Self-critique (if necessary):

4 x y^3 + 6 x^2 y^2 y' - 3 y^2 - 6xy y' = 0 I understand it down to this step. I don’t understand why the y’ was left in on those 2 terms.It looks like everything was used in determining the derivative of the entire expression. I did not work through the rest of it because I thought that that part would be pretty straightforward.

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Self-critique Rating: 2

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Question: `q008. (Mth 173 only; Mth 271 doesn't require the use of sine and cosine functions). Follow the procedure of the preceding problem to determine the value of y ' when x = 3 and y = `pi, for the equation x^2 sin (y) - sin(xy) = 0.

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Your solution:

Mth 271

confidence rating #$&*:

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Given Solution:

`aTaking the derivative of both sides of the equation we obtain

(x^2) ' sin(y) + x^2 (sin(y) ) ' - ( sin (xy) ) '.

By the Chain Rule

(sin(y)) ' = y ' cos(y) and

(sin(xy)) ' = (xy) ' cos(xy) = ( x ' y + x y ' ) cos(xy) = (y + x y ' ) cos(xy).

So the derivative of the equation becomes

2 x sin(y) + x^2 ( y ' cos(y)) - ( y + x y ' ) cos(xy) = 0. Expanding all terms we get

2 x sin(y) + x^2 cos(y) y ' - y cos(xy) - x cos(xy) y ' = 0. Leaving the y ' terms on the left and factoring ou y ' gives us

[ x^2 cos(y) - x cos(xy) ] y ' = y cos(xy) - 2x sin(y), so that

y ' = [ y cos(xy) - 2x sin(y) ] / [ x^2 cos(y) - x cos(xy) ].

Now we can substitute x = 3 and y = `pi to get

y ' = [ `pi cos( 3 * `pi) - 2 * 3 sin(`pi) ] / [ 3^2 cos(`pi) - 3 cos(3 * `pi) ] = [ `pi * -1 - 2 * 3 * 0 ] / [ 9 * -1 - 3 * -1 ] = -`pi / 6.

STUDENT QUESTION

I think that the derivative of sin (xy) is cos (xy) but in the solution ( y + x y ' ) cos(xy) is the derivitive. I had this

same problem on a earlier problem also, can you maybe explain to me were I am messing this up?

Or were the y+xy` came from?

INSTRUCTOR RESPONSE

The derivative of sin(xy) is (xy)' * cos(xy), by the chain rule.

(xy)' = x ' y + x y ' = 1 * y + x y ' = y + x y '.

Thus the derivative of sin(xy) is (y + x y ' ) cos(xy).

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Self-critique (if necessary):

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&#Good responses. See my notes and let me know if you have questions. &#