mat01a1: implicit differentiation · implicit di erentiation: up until now we have only studied...

MAT01A1: Implicit Differentiation

Dr Craig

Week: 4 May 2020

We emphasized when we covered Ch 3.4 that

the Chain Rule was a very important topic.

Let us re-cap that with two examples before

we start with the new differentiation

techniques.

Reminder: the Chain Rule

If f and g are both differentiable and

F = f ◦ g is the composite function

defined by

F (x) = f (g(x)),

then F is differentiable and F ′ is given by

F ′(x) = f ′(g(x)).g′(x)

Chain rule examples

Find y′ for each of the following functions

I y = e2x cot( 3√x)

I y = 2etan(x2)

Try these on your own first. Solutions are on

the slides that follow.

Chain rule examples: solutions

(e2x cot( 3

√x))

=(cot( 3√x)) ddx

(e2x)+(e2x) ddx

(cot( 3√x))

=(cot( 3√x))e2x

dx(2x) +

(e2x) (− csc2( 3

√x)) ddx

( 3√x)

=(cot( 3√x))e2x(2) +

(e2x) (− csc2( 3

√x)) 1

33√x2

= e2x(2 cot( 3

√x)− csc2( 3

33√x2

Chain rule examples: solutions

tan(x2))

tan(x2))(ln 2)

(etan(x

tan(x2))(ln 2)

(etan(x

(tan(x2)

tan(x2))(ln 2)

(etan(x

sec2(x2)) ddx

tan(x2))(ln 2)

(etan(x

sec2(x2))2x

Implicit differentiation:

Up until now we have only studied functions

where one variable (usually y) is defined in

terms of another variable (usually x).

Now we will consider relations between x

and y values where the two variables are

defined in terms of one another. For

example:

x2 + y2 = 25

Note: this curve is not a function as it does

not pass the vertical line test.

defined in terms of one another.

example:

x2 + y2 = 25

defined in terms of one another. For

example:

x2 + y2 = 25

Clearly this curve is not a function

x2 + y2 = 25

We want a method of calculating the slope

of the tangent to the curve x2 + y2 = 25.

We could try to solve for y, but then we get

y = ±√

25− x2

Thus the implicit relation x2 + y2 = 25 can

only be fully described by the twofunctions below:

f (x) =√

25− x2 & g(x) = −√

25− x2

We see this in the sketches on the next slide.

y = ±√

25− x2

f (x) =√

25− x2 & g(x) = −√

25− x2

y = ±√

25− x2

f (x) =√

25− x2 & g(x) = −√

25− x2

In pictures...

f (x) =√25− x2 and g(x) = −

√25− x2

We could use existing techniques and these

curves to find the slope of a tangent.

Fortunately there is a much better method.

In pictures...

f (x) =√25− x2 and g(x) = −

√25− x2

We could use existing techniques and these

curves to find the slope of a tangent.

Fortunately there is a much better method.

We are interested in differentiating relations

which are implicitly defined.

To do this

we treat y as a function of x

and then apply our existing rules of

differentiation.

A crucial rule will be the

chain rule. For example:

dx(y2) = 2y.

dx(y) = 2y.y′

To do this

differentiation. A crucial rule will be the

chain rule.

For example:

dx(y2) = 2y.

dx(y) = 2y.y′

To do this

differentiation. A crucial rule will be the

chain rule. For example:

dx(y2) = 2y.

dx(y) = 2y.y′

We apply this approach to our first example:

(x2 + y2

dx(25)

dx(x2) +

dx(y2) = 0

∴ 2x + 2y.y′ = 0

∴ y′ =−2x2y

=−xy

Note that y′ depends on BOTH x and y so

we can’t talk about the derivative at x = a.

We apply this approach to our first example:d

(x2 + y2

dx(25)

dx(x2) +

dx(y2) = 0

∴ 2x + 2y.y′ = 0

∴ y′ =−2x2y

=−xy

(x2 + y2

dx(25)

dx(x2) +

dx(y2) = 0

∴ 2x + 2y.y′ = 0

∴ y′ =−2x2y

=−xy

(x2 + y2

dx(25)

dx(x2) +

dx(y2) = 0

∴ 2x + 2y.y′ = 0

∴ y′ =−2x2y

=−xy

(x2 + y2

dx(25)

dx(x2) +

dx(y2) = 0

∴ 2x + 2y.y′ = 0

∴ y′ =−2x2y

(x2 + y2

dx(25)

dx(x2) +

dx(y2) = 0

∴ 2x + 2y.y′ = 0

∴ y′ =−2x2y

=−xy

(x2 + y2

dx(25)

dx(x2) +

dx(y2) = 0

∴ 2x + 2y.y′ = 0

∴ y′ =−2x2y

=−xy

(x2 + y2

dx(25)

dx(x2) +

dx(y2) = 0

∴ 2x + 2y.y′ = 0

∴ y′ =−2x2y

=−xy

However, we can calculate y′ at a point by

using both the x and y coordinates.

The slope of a tangent at a point

For y2 + x2 = 25 we have y′ =−xy

We emphasize that we need both the x and

y coordinate of a point to calculate y′ at that

point. Consider the points (3, 4) and (3,−4).

We get m = −34 for the gradient of the

tangent to the curve at the first point.

Substituting into y = mx + c we get

y = −0.75x + 6.25 as an equation of the

tangent.

point.

Consider the points (3, 4) and (3,−4).

tangent.

An equation for the tangent at the point

(3, 4) is y = −0.75x + 6.25.

An equation for the tangent at the point

(3,−4) is y = 0.75x− 6.25.

The Folium of Descartes

x3 + y3 = (3a)xy

Huh? What is a folium?

The Folium of Descartes continued...

x3 + y3 = (3a)xy

Let us consider the case of a = 2. Now use

implicit differentiation to find y′ of

x3 + y3 = 6xy

Find the equation of the tangent at the point

(3, 3).

We differentiate:

(x3 + y3

dx(6xy)

Remember: we regard y as function of x.

So, on the RHS we think of 6xy as the

product of 6x and y.

∴ 3x2 + 3y2.y′ =d

dx(6x)y +

dx(y)(6x)

∴ 3x2 + 3y2.y′ = 6y + y′6x

Now we solve for y′.

We differentiate:

(x3 + y3

dx(6xy)

∴ 3x2 + 3y2.y′ =d

dx(6x)y +

dx(y)(6x)

∴ 3x2 + 3y2.y′ = 6y + y′6x

We differentiate:

(x3 + y3

dx(6xy)

∴ 3x2 + 3y2.y′ =d

dx(6x)y +

dx(y)(6x)

∴ 3x2 + 3y2.y′ = 6y + y′6x

We differentiate:

(x3 + y3

dx(6xy)

∴ 3x2 + 3y2.y′ =d

dx(6x)y +

dx(y)(6x)

∴ 3x2 + 3y2.y′ = 6y + y′6x

We differentiate:

(x3 + y3

dx(6xy)

∴ 3x2 + 3y2.y′ =d

dx(6x)y +

dx(y)(6x)

∴ 3x2 + 3y2.y′ = 6y + y′6x

We had: 3x2 + 3y2.y′ = 6y + y′6x

∴ 3y2.y′ − y′6x = 6y − 3x2

y′ =6y − 3x2

3y2 − 6x=

2y − x2

y2 − 2x

Now, at the point (3, 3) we get y′ = −1.

Substituting into y = −x + c we get

y = −x + 6.

We had: 3x2 + 3y2.y′ = 6y + y′6x

∴ 3y2.y′ − y′6x = 6y − 3x2

y′ =6y − 3x2

3y2 − 6x=

2y − x2

y2 − 2x

y = −x + 6.

We had: 3x2 + 3y2.y′ = 6y + y′6x

∴ 3y2.y′ − y′6x = 6y − 3x2

y′ =6y − 3x2

3y2 − 6x=

2y − x2

y2 − 2x

y = −x + 6.

We had: 3x2 + 3y2.y′ = 6y + y′6x

∴ 3y2.y′ − y′6x = 6y − 3x2

y′ =6y − 3x2

3y2 − 6x=

2y − x2

y2 − 2x

y = −x + 6.

We had: 3x2 + 3y2.y′ = 6y + y′6x

∴ 3y2.y′ − y′6x = 6y − 3x2

y′ =6y − 3x2

3y2 − 6x=

2y − x2

y2 − 2x

y = −x + 6.

Examples:

1. Find y′ if sin(x + y) = y2 cosx

2. Find y′′ if x4 + y4 = 16

Remember that we consider y to be a

function of x.

Try each of these and then look at the

solutions on the slides that follow.

A big hint for (2.) is that you should

substitute y′ and the original equation into

your calculation for y′′.

Example: sin(x + y) = y2 cosx

Solution:d

dx(sin(x + y)) =

dx(y2 cosx).

We must apply the Chain Rule to the LHS

and treat the RHS as a product. So,

cos(x + y)d

dx(x + y)=

dx(y2)(cosx) +

dx(cosx)y2

∴ cos(x+y)(1+y′) = 2yy′ cosx−(sinx)y2

∴ cos(x + y) + y′ cos(x + y) = 2yy′ cosx− y2 sinx

∴ y′ cos(x+ y)− y′2y cosx=−y2 sinx− cos(x+ y)

Solution:d

dx(sin(x + y)) =

dx(y2 cosx).

cos(x + y)d

dx(x + y)=

dx(y2)(cosx) +

dx(cosx)y2

Solution:d

dx(sin(x + y)) =

dx(y2 cosx).

cos(x + y)d

dx(x + y)=

dx(y2)(cosx) +

dx(cosx)y2

Solution:d

dx(sin(x + y)) =

dx(y2 cosx).

cos(x + y)d

dx(x + y)=

dx(y2)(cosx) +

dx(cosx)y2

Solution:d

dx(sin(x + y)) =

dx(y2 cosx).

cos(x + y)d

dx(x + y)=

dx(y2)(cosx) +

dx(cosx)y2

We had

y′ cos(x+y)−y′2y cosx = −y2 sinx−cos(x+y)

Taking out y′ as a common factor on the

LHS gives:

y′(cos(x+y)−2y cosx) = −y2 sinx−cos(x+y)

Finally we get:

y′ =y2 sinx + cos(x + y)

2y cosx− cos(x + y)

We had

LHS gives:

Finally we get:

We had

LHS gives:

Finally we get:

Solution:d

(x4 + y4

dx(16), so

4x3 + 4y3y′ = 0 and hence y′ =−x3

We apply the quotient rule to get:

y′′ = −ddx(x

3)y3 − x3 ddx(y3)

∴ y′′ = −3x2y3 − x3(3y2y′)

Now we substitute y′ back in . . .

Solution:d

(x4 + y4

dx(16), so

4x3 + 4y3y′ = 0

and hence y′ =−x3

y′′ = −ddx(x

3)y3 − x3 ddx(y3)

∴ y′′ = −3x2y3 − x3(3y2y′)

Solution:d

(x4 + y4

dx(16), so

4x3 + 4y3y′ = 0 and hence y′ =−x3

y′′ = −ddx(x

3)y3 − x3 ddx(y3)

∴ y′′ = −3x2y3 − x3(3y2y′)

Solution:d

(x4 + y4

dx(16), so

4x3 + 4y3y′ = 0 and hence y′ =−x3

y′′ = −ddx(x

3)y3 − x3 ddx(y3)

∴ y′′ = −3x2y3 − x3(3y2y′)

Solution:d

(x4 + y4

dx(16), so

4x3 + 4y3y′ = 0 and hence y′ =−x3

y′′ = −ddx(x

3)y3 − x3 ddx(y3)

∴ y′′ = −3x2y3 − x3(3y2y′)

∴ y′′ = −3x2y3 − 3x3y2−x

Simplifying gives us

y′′ = −3(x2y4 + x6)

y7= −3x

2(y4 + x4)

Now we substitute the original equation to

y′′ = −3x2(16)

y7=−48x2

∴ y′′ = −3x2y3 − 3x3y2−x

y′′ = −3(x2y4 + x6)

y7= −3x

2(y4 + x4)

y′′ = −3x2(16)

y7=−48x2

∴ y′′ = −3x2y3 − 3x3y2−x

y′′ = −3(x2y4 + x6)

y7= −3x

2(y4 + x4)

y′′ = −3x2(16)

y7=−48x2

Differentiating inverse trig functions:

The calculation of the derivatives of

y = arcsinx and y = arctanx are shown in

the textbook.

Here we show how to get the derivative of

y = cos−1(x)

Remember that inverse trig functions are

functions. Therefore we should expect to be

able to get their derivatives as functions of x.

The calculation of the derivatives of

y = arcsinx and y = arctanx are shown in

the textbook.

Here we show how to get the derivative of

y = cos−1(x)

Remember that inverse trig functions are

functions. Therefore we should expect to be

able to get their derivatives as functions of x.

Recall that

y = cos−1(x) ⇐⇒ x = cos y and 0 6 y 6 π

Instead of trying to differentiate

y = cos−1(x), we rather apply implicit

differentiation to cos y = x.

dxcos y =

∴ (− sin y)(y′) = 1 ⇒ y′ = −1/(sin y)

But, we want to express y′ in terms of x.

Recall that

dxcos y =

∴ (− sin y)(y′) = 1 ⇒ y′ = −1/(sin y)

Recall that

dxcos y =

∴ (− sin y)(y′) = 1 ⇒ y′ = −1/(sin y)

Recall that

dxcos y =

∴ (− sin y)(y′) = 1

⇒ y′ = −1/(sin y)

Recall that

dxcos y =

∴ (− sin y)(y′) = 1 ⇒ y′ = −1/(sin y)

Recall that

dxcos y =

∴ (− sin y)(y′) = 1 ⇒ y′ = −1/(sin y)

Differentiating y = cos−1(x) continued. . .

Since cos y = x we get cos2(y) = x2.

∴ 1−sin2(y) = x2 (cos2 θ+sin2 θ = 1)

∴ 1− x2 = sin2(y)

∴√

1− x2 =√

sin2 y

∴√

1− x2 = | sin y| = sin y (0 6 y 6 π)

Finally: y′ =−1√1− x2

∴ 1−sin2(y) = x2 (cos2 θ+sin2 θ = 1)

∴ 1− x2 = sin2(y)

∴√

1− x2 =√

sin2 y

∴√

1− x2 = | sin y| = sin y (0 6 y 6 π)

∴ 1−sin2(y) = x2 (cos2 θ+sin2 θ = 1)

∴ 1− x2 = sin2(y)

∴√

1− x2 =√

sin2 y

∴√

1− x2 = | sin y| = sin y (0 6 y 6 π)

∴ 1−sin2(y) = x2 (cos2 θ+sin2 θ = 1)

∴ 1− x2 = sin2(y)

∴√

1− x2 =√

sin2 y

∴√

1− x2 = | sin y| = sin y (0 6 y 6 π)

∴ 1−sin2(y) = x2 (cos2 θ+sin2 θ = 1)

∴ 1− x2 = sin2(y)

∴√

1− x2 =√

sin2 y

∴√1− x2 = | sin y| = sin y (0 6 y 6 π)

∴ 1−sin2(y) = x2 (cos2 θ+sin2 θ = 1)

∴ 1− x2 = sin2(y)

∴√

1− x2 =√

sin2 y

∴√1− x2 = | sin y| = sin y (0 6 y 6 π)

Note: the previous slide presents one way of

getting sin y in terms of x.

Since cos y = x =x

1, we could also draw a

triangle with an angle of y and Hyp. = 1,

Adj. = x.

That would give us Opp. =√1− x2.

Therefore

sin y =√1− x2.

Derivatives of inverse trig functions:

dx(sin−1 x) =

1√1− x2

dx(csc−1 x) =

−1x√x2 − 1

dx(cos−1 x) =

−1√1− x2

dx(sec−1 x) =

x√x2 − 1

dx(tan−1 x) =

1 + x2d

dx(cot−1 x) =

−11 + x2

Comparing y = arctanx and its derivative

y = tan−1(x) andd

(tan−1 x

Example: Differentiate y = x arctan√x.

Solution:

dx(x) arctan

√x + x

(arctan

= arctan√x + x

1 + (√x)2

)· ddx

= arctan√x + x

= arctan√x +

2(1 + x)

Solution:

dx(x) arctan

√x + x

(arctan

= arctan√x + x

1 + (√x)2

)· ddx

= arctan√x + x

= arctan√x +

2(1 + x)

Solution:

dx(x) arctan

√x + x

(arctan

= arctan√x + x

1 + (√x)2

)· ddx

= arctan√x + x

= arctan√x +

2(1 + x)

Solution:

dx(x) arctan

√x + x

(arctan

= arctan√x + x

1 + (√x)2

)· ddx

= arctan√x + x

= arctan√x +

2(1 + x)

Solution:

dx(x) arctan

√x + x

(arctan

= arctan√x + x

1 + (√x)2

)· ddx

= arctan√x + x

= arctan√x +

2(1 + x)

The prescribed tut problems for this section

are on the next slide. You should complete

them before going through the slides for

Ch 3.6.

But first, here is an interesting story about

Rene Descartes (remember our second

example of implicit differentiation used the

Folium of Descartes):

https://wild.maths.org/rene-descartes-and-

fly-ceiling

Prescribed tut problems:

Complete the following exercises from the

8th edition:

Ch 3.5:

1, 3, 5, 7, 11, 17, 19, 20, 25, 27, 35, 37, 49,

53, 60

mat01a1: implicit differentiation · implicit di erentiation: up until now we have only studied...

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