query assn 19

course Mth 173

assignment #019£l±Ý‘ñØ¦‰’íÀUxÍ´x‚óq ê‹T{©àÅ¹§

Calculus I

11-03-2007

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22:35:14

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RESPONSE -->

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22:54:09

Query problem 3.4.29 (3d edition 3.4.20) was 4.4.12 Derivative of `sqrt( (x^2*5^x)^3

What is the derivative of the given function?

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RESPONSE -->

y= ''sqrt(x^2 * 5^x)^3

y= ((x^2 * 5^x)^3)^.5

y'=.5(3(2x * 5^x + ln(5) 5^x *x^2)^2)^-.5

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23:03:04

** The function is `sqrt( (x^2 * 5^x)^3 ) = (x^2 * 5^x)^(3/2).

This is of form f(g(x)) with g(x) = x^2 * 5^x and f(z) = z^(3/2). Thus when you substitute you get f(g(x)) = g(x)^(3/2) = (x^2 * 5^x)^(3/2).

(x^2 * 5^x) ' = (x^2)' * 5^x + x^2 * (5^x) ' =

2x * 5^x + x^2 ln 5 * 5^x =

(2x + x^2 ln 5) * 5^x.

`sqrt(z^3) = z^(3/2), so using w(x) = f(g(x)) with f(z) = z^(3/2) and g(x) = x^2 * 5^x we get

w ' = (2x + x^2 ln 5) * 5^x * [3/2 (x^2 * 5^x)^(1/2)] = 3/2 (2x + x^2 ln 5) * | x | * 5^(3/2 x).

Note that sqrt(x^2) is | x |, not just x, since the square root must be positive and x might not be. **

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RESPONSE -->

by breaking the equation down into it's seperate parts of z = (x^2 * 5^x) and f(z) = z^3/2 we can use the product rule to find the derivative of z and the chain rule to link it to f(z)

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23:07:54

Query problem 3.4.28 (3d edition 3.4.19) (was 4.4.20) derivative of 2^(5t-1).

What is the derivative of the given function?

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RESPONSE -->

y=2^(5t-1)

f(g(t))= 2^(5t-1)

f(g) = 2^g

g(t)= 5t-1

y'= f'(g(t))

=ln(g) 2^g

=ln(5t-1) * 2^(5t-1)

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23:11:28

This function is a composite. The inner function is g(x)=5t-1 and the outer function is f(z)=2^z.

f'(z)=ln(2) * 2^z.

g ' (x)=5

(f(g(t)) ' = g ' (t)f ' (g(t))=

5 ln(2) * 2^(5t-1).

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RESPONSE -->

the dirivative of the inner function is 5 so that becomes the outside function in the derivative.

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23:33:24

**** Query 3.4.68 (3d edition 3.4.56) y = k (x), y ' (1) = 2.

What is the derivative of k(2x) when x = 1/2?

What is the derivative of k(x+1) when x = 0?

{]What is the derivative of k(x/4) when x = 4?

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RESPONSE -->

y=k(2x)

f(g(x)) = k(2x)

f(g(x))' = g'(x)f'(g(x))

f(g) = k *g

g(x) = 2x

g'(x)f'(g(x)) = 2 * (k * (2x) + 2 *k)

= 2(3k(2x))

y'(1/2)= 6k

f(g(x))= k(x+1)

f(g(x))'= x * (k * (x+1) + x *k))

f(g(0)'= 0

f(g(x))= k(x/4)

f(g) = k(g)

g(x)= .25 * x

f(g(x))= .25* (k(.25 x))

f(g(4))= k(.25)

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23:36:56

** We apply the Chain Rule:

( k(2x) ) ' = (2x) ' * k'(2x) = 2 k(2x).

When x = 1/2 we have 2x = 1.

k ' (1) = y ' (1) = 2 so

when x = 1/2

( k(2x) ) ' = 2 k(2 * 1/2) = 2 * k'(1) = 2 * 2 = 4.

(k(x+1)) ' = (x+1)' k ' (x+1) = k ' (x+1) so

when x = 0 we have

(k(x+1) ) ' = k ' (x+1) = k ' (1) = 2

(k(x/4)) ' = (x/4)' k'(x/4) = 1/4 * k'(x/4) so when x = 4 we have

(k(x/4))' = 1/4 * k'(x/4) = 1/4 k'(4/4) = 1/4 * 2 = 1/2. **

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RESPONSE -->

okay

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23:44:22

Query 3.4.81 (3d edition 3.4.68). Q = Q0 e^(-t/(RC)). I = dQ/dt.

Show that Q(t) and I(t) both have the same time constant.

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RESPONSE -->

Q('dt)= Q0 e^(-t/(RC))

l = 'dQ / 'dt

we can see that the time constant is 'dt for both equations

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23:54:21

** We use the Chain Rule.

(e^(-t/(RC)))' = (-t/(RC))' * e^(-t/(RC)) = -1/(RC) * e^(-t/(RC)).

So dQ/dt = -Q0/(RC) * e^(-t/(RC)).

Both functions are equal to a constant factor multiplied by e^(-t/(RC)).

The time constant for both functions is therefore identical, and equal to RC. **

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RESPONSE -->

I didn't follow this at all, I can see the application of the chain rule and that was about all I could see in the answer, could you please explain this further.

dQ/dt means the derivative of Q with respect to t.

If we write Q ' (t) it is because we know that the variable is t and it is implicit that the derivative will be with respect to t.

The notation dQ/dt is explicit about what the function is and with respect to what variable the derivative is being taken.

dQ/dt = Q ' (t) = -Q0 / (R C) * e^(-t / (R C)). This is by the Chain Rule with g(t) = -t / (R C) and f(z) = e^z.

The time constant is the time required for the exponential factor to decrease by a factor of e, e.g., from e^0 = 0 to e^-1. In this case the time constant is RC, because in time interval R C the function decreases by factor e.

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23:59:38

Query problem 3.5.5 (unchanged since 3d edition) (formerly 4.5.6). What is the derivative of sin(3x)

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RESPONSE -->

f(g(x))= sin(3x)

f(g)= sin(g)

g(x)= 3x

f(g(x)' = g'(x) * f'(g(x))

= 3 * cos(3x)

=3cos(3x)

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23:59:54

** sin(3x) is the composite of g(x) = 3x, which is the 'inner' function (the first function that operates on the variable x) and the 'outer' function f(z) = sin(z).

Thus f(g(x)) = sin(g(x)) = sin(3x).

The derivative is (f (g(x) ) ' = g ' (x) * f' ( g(x) ).

g ' (x) = (3x) ' = 3 * x ' = 3 ', and

f ' (z) = (sin(z) ) ' = cos(z).

So the derivative is [ sin(3x) ] ' = ( f(g(x) ) ' = g ' (x) * f ' (g(x) ) = 3 * cos(3x). **

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RESPONSE -->

okay

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00:16:50

Query problem 3.5.48 (3d edition 3.5.50) (formerly 4.5.36). Give the equations of the tangent lines to graph of y = sin(x) at x = 0 and at `pi/3

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RESPONSE -->

y= sin(x)

y'= cos(x)

x y y'

0, 0, 1

'pi/3, .866 .5

since y'(0) = 1 so we get

0= 1(0)+b

b=0

tan(0)= 1x +0

since y'(0)= .5 we get

.866= .5(1.047) + b

.866= .5236 +b

b= .3424

tan('pi/3)= .5x + .3424

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00:20:53

** At x = 0 we have y = 0 and y ' = cos(0) = 1.

The tangent line is therefore the line with slope 1 through (0,0), so the line is y - 0 = 1 ( x - 0) or just y = x.

At x = `pi/3 we have y = sin(`pi/3) = `sqrt(3) / 2 and y ' = cos(`pi/3) = .5.

Thus the tangent line has slope .5 and passes thru (`pi/3,`sqrt(3)/2), so its equation is

y - `sqrt(3)/2 = .5 (x - `pi/3)

y = .5 x - `pi/6 + `sqrt(3)/2. Approximating:

y - .87 = .5 x - .52. So

y = .5 x + .25, approx.

Our approximation to sin(`pi/6), based on the first tangent line:

The first tangent line is y = x. So the approximation at x = `pi / 6 is

y = `pi / 6 = 3.14 / 6 = .52, approximately.

Our approximation to sin(`pi/6), based on the second tangent line, is:

y = .5 * .52 + .34 = .60.

`pi/6 is equidistant from x=0 and x=`pi/3, so we might expect the accuracy to be the same whichever point we use.

The actual value of sin(`pi/6) is .5. The approximation based on the tangent line at x = 0 is .52, which is much closer to .5 than the .60 based on the tangent line at x = `pi/3.

The reason for this isn't too difficult to see. The slope is changing more quickly around x = `pi/3 than around x = 0. Thus the tangent line will more more rapidly away from the actual function near x = `pi/3 than near x = `pi/2. **

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RESPONSE -->

I didn't use the 'pi notation but other than that our answers are similiar.

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00:32:16

Query 3.5.34 (3d edition 3.5.40). Der of sin(sin x + cos x)

What is the derivative of the given function and how did you find it?

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RESPONSE -->

y= sin(sin(x) + cos(x))

f(g(x))= sin(sin(x) + cos(x))

f(g)= sin(g)

g(x)= sin(x) + cos(x)

g(x)'= cos(x) * cos(x) + -sin(x) * sin(x)

g(x)'= cos(x)^2 + -sin(x) * sin(x)

f(g(x)'= g'(x) * f'(g(x)

= (cos(x)^2 + -sin(x) * sin(x)) * cos(sin(x) + cos(x))

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00:33:10

The function y = sin( sin(x) + cos(x) ) is the composite of g(x) = sin(x) + cos(x) and f(z) = sin(z).

The derivative of the composite is g ' (x) * f ' (g(x) ).

g ' (x) = (sin(x) + cos(x) ) ' = cos(x) - sin(x).

f ' (z) = sin(z) ' = cos(z).

So g ' (x) * f ' (g(x)) = ( cos(x) - sin(x) ) * cos( sin(x) + cos(x) ).

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RESPONSE -->

okay

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00:36:53

Query Add comments on any surprises or insights you experienced as a result of this assignment.

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RESPONSE -->

I think that I find the basic concept of the chain rule and I can find the derivative of the basic elements of the more complex equations, but when ""chaining"" these together I am having a little trouble in the simplification of the entire process. I think that it will take a while to sink in.

The 'chaining' idea is easiest to represent using the dy/dx notation.

For example if we have the composite f(g(x)) we want to find df/dx, meaning (f(g(x)) ' where the ' indicates the derivative with respect to x.
The 'chain' is the function f(z) applied to the function z = g(x). The f ' (z) function is written df/dz, and the g'(x) function is dz/dx. So

f ' (x) = g ' (x) * f ' (g(x)) can be written as
df/dx = dz/dx * df/dz, which we rewrite using the commutativity of multiplication in the 'chaining' order
df/dx = df/dz * dz/dx. Note that if the derivatives were algebraic expressions (which they are not) the dz on the right-hand side would 'cancel' leaving df/dx.
We write this with the understanding that we will in the end substitute the g(x) function for z.

For example f(x) = cos(x^2) is the composite of f(z) = cos(z) with z = x^2. Then since df/dz = (cos(z)) ' = - sin(z) and dz/dx = (x^2) ' = 2 x we have

df / dx = df/dz * dz/dx = -sin(z) * 2x. We substitute x^2 for z to obtain
df/dx = -sin(x^2) * 2x, which we then write in the standard form
df/dx = - 2 x sin(x^2).

Every part of this can be understood in the f ' (g(x)) format with f(z) = cos(z) and g(x) = x^2. We get f ' (x) = g ' (x) f ' (g(x)) = 2x * (-sin(x^2)) = - 2 x sin(x^2), the same result we obtained in the 'chaining' notation.

We can actually write f(x) as a 'chain' of any length. For example we could have a 'chain' consisting of f(z), z(u), u(w) and w(x). The derivative would be

df/dx = df/dz * dz/du * du/dw * dw/dx.

The 'chaining' notation proves its value when we have a multiple composite like e^(sin(x^2)). We can write this as f(g(x)), with f(z) = e^z and g(x) = sin(x^2), but the g(x) function is still a composite. Or we could write f(z) = e^sin(z) and g(x) = x^2, and the f function would be a composite. We can deal with that easily enough, but we're stuck with needing another application of the chain rule (either to g ' (x) or f ' (z) ), and that can get confusing.

If we write the function e^(sin(x^2)) as the 'chain' f(z) = e^z, z(u) = sin(u), u(x) = x^2, then we will calculate df/dz = e^z, dz/du = cos(u) and du/dx = 2 x. Our derivative will be

df/dx = df/dz * dz/du * du/dx
= e^z * cos(u) * 2x
= e^(sin(u)) cos(x^2) * 2 x
= e^(sin(x^2)) * cos(x^2) * 2x
= 2x cos(x^2) e^(sin(x^2)).

Had we written the original function as f(g(x)) with f(z) = e^z and g(x) = sin(x^2) we could have easily enough found that f ' (z) = e^z and g ' (x) = 2 x cos(x^2), so that (f (g(x)) ' = 2 x cos(x^2) e^(sin(x^2)). This agrees, as it must, with the result we just obtained by 'chaining'. However note that to calculate g ' (x) we had to use the chain rule a second time.

If we had a more complex function like ln ( sin(e^(cos(x^2)))), which is a composite of five different functions, the f (g(x)) breakdowns would start to get complicated. However the 'chain' f(z) = ln(z), z(u) = sin(u), u(w) = e^w, w(r) = cos(r) and finally r(x) = x^2 is much easier to deal with. We get

df/dx = df/dz * dz/du * du/dw * dw/dr * dr/dx = 1 / z * cos(u) * e^w * (-sin(r) ) * 2x. It takes a few steps to back-substitute but this isn't difficult:

1 / z * cos(u) * e^w * (-sin(r) ) * 2x
= -1 / sin(u) * cos(u) * e^w * sin(r) * 2x
= - 1 / sin(e^w) * cos(e^w) * e^w * sin(r) * 2x
= - 1 / sin(e^(cos(r))) * cos(e^(cos(r))) * e^(cos(r)) * sin(r) * 2x
= - 1 / sin(e^(cos(x^2))) * cos(e^(cos(x^2))) * e^(cos(x^2)) * sin(x^2) * 2x

The substitution proceeds one function at a time, from left to right.
In final form we would reverse the order of terms to obtain

-2x sin(x^2) e^(cos(x^2))) * cos(e^(cos(x^2))) / sin(e^(cos(x^2))).

Incidentally this simplifies further to -2x sin(x^2) e^(cos(x^2))) * cotan(e^(cos(x^2))).

The names of the 'intermediate' variables z, u, w, and r are completely arbitrary. These variables appear in order to do the intermediate steps, then disappear again without a trace as we complete our 'back-substitutions'. They are called 'dummy' variables.

We can introduce as many 'dummy' variables as we require in order to make our 'chain'.

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query assn 19

Good work.

See my notes on 'chaining', just above this note

&#

Let me know if you have questions. &#

query assn 19

Good work. See my notes on 'chaining', just above this note

&#

Let me know if you have questions. &#

Good work.

See my notes on 'chaining', just above this note