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#$&*

course MTH272

If your solution to a stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution: If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

003. `query 3

Note that the documents listed below are recommended as a summary, in q_a_ form, of the rules and applications of differentiation. You are welcome to submit any or all of these, or any part of any of these documents. They are recommended for anyone who isn't completely sure of the present chapter.

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa11.htm

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa12.htm

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa13.htm

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa14.htm

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa15.htm

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa16.htm

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Question: `q 4.5.5 (previously 4.5.10 (was 4.4.10)) find the derivative of ln(1-x)^(1/3)

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Your solution:

ln(1-x)^(1/3)=d/dx 1/3 ln(1-x) u=1-x

du/dx = 1/3 ln(1-x)`* u using the chain rule

du/dx = 1/3 ln(1-x)`* 1- x 1- x^1 ### 1 goes to 0 ### -x^1 = -1

1/3 ln (1-x)*-1= (1/3) ( -1/(1-x ) )

confidence rating #$&*:2

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Given Solution:

Note that ln(1-x)^(1/3) = 1/3 ln(1-x)

The derivative of ln(1-x) is u ' * 1/u with u = 1-x. It follows that u ' = -1 so the derivative of ln(1-x) is -1 * 1/(1-x) = -1/(1-x).

The derivative of 1/3 ln(1-x) is therefore 1/3 * -1/(1-x) = -1 / [ 3(1-x) ].**

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Self-critique (if necessary):I guess I missed a step by not multiplying 1/3 into the rest of function.

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Self-critique Rating:3

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Question: `q4.5.9 (previously 4.5.25 (was 4.4.24)) find the derivative of ln( (e^x + e^-x) / 2)

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Your solution:

ln( (e^x + e^-x) / 2) = (du/dx) (ln( (e^x + e^-x) / 2) u = ln (e^x + e^-x) ) = 1/u the derivative of the original function multiplied by u

(du/dx) = (e^x - e^-x) the 2 was a constant went to 0, the negative exponent transfers to e

u = (e^x + e^-x)

(e^x - e^-x) (1/(e^x + e^-x) )

{(e^x - e^-x) / (e^x + e^-x)}

confidence rating #$&*:2

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Given Solution:

`a the derivative of ln(u) is 1/u du/dx; u = (e^x + e^-x)/2 so du/dx = (e^x - e^-x) / 2.

The term - e^(-x) came from applying the chain rule to e^-x.

The derivative of ln( (e^x + e^-x) / 2) is therefore

[(e^x - e^-2) / 2 ] / ] [ (e^x + e^-x) / 2 ] = (e^x - e^-x) / (e^x + e^-x).

This expression does not simplify, though it can be expressed in various forms (e.g., (1 - e^-(2x) ) / ( 1 + e^-(2x) ), obtained by dividing both numerator and denominator by e^x.).

ALTERNATIVE SOLUTION:

ln( (e^x + e^-x) / 2) = ln( (e^x + e^-x) ) - ln(2).

the derivative of e^(-x) is - e^(-x) and ln(2) is constant so its derivative is zero.

So you get

y ' = (e^x - e^-x)/(e^x + e^-x). **

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Self-critique (if necessary):n/a

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Self-critique Rating:

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Question: `q 4.5.10 (previously 4.5.30 (was 4.4.30) ) write log{base 3}(x) with base e

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Your solution:

Log3x = logarithm of the base

confidence rating #$&*:3

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Given Solution:

`a

We know that

log{base b}(a) = log(a) / log(b), so that

log(a) = log(b) * log{base b}(a),

where the 'log' in log(a) and log(b) can stand for the logarithm to any base.

In particular, if this 'log' stands for the base-e logarithm, we write it as 'ln' (which stands for 'natural log'), and we could write the above as

log{base b}(a) = ln(a) / ln(b).

The expression in the current problem can therefore be written as

log{base 3}(x) = ln(x) / ln(3).

It's worth noting also that

y = log{base 3}(x) means that x = 3^y; i.e., y is the power to which 3 must be raised to give us x.

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Self-critique (if necessary):

I understood the principle that log{base 3}(x) = ln(x) / ln(3) but is this the way that you wanted the problem answered. The way I understood it was to write just log{base 3}(x) or xlog3

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Self-critique Rating:3

@& The rule is that log(base b)(x) = log(x) / log(b).

This rule follows directly from the definition of the log{base b} funciton as the inverse of the funciton y = b^x. It's very desirable to understand this connection, but it's necessary to simply know the above-stated rule.*@

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Question: `q4.5.22 (previously extra prob (was 4.4.50)). Find the equation of the line tangent to the graph of 25^(2x^2) at (-1/2,5)

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Your solution:

25^(2x^2) at the point (-1/2,5) x=-1/2

u=2x^2 4x get the derivative of u

[(4x)ln(25)][25^(2x^2)] rewrite the function

[(4(-1/2)(ln25)][25^(2(-1/2)^2)]= using the chain rule multiply the two

{-2(ln25)}{25^-.5}=-1.288 multiply it out

(y - 5) = -1.288( x - -1/2) Using (y - y1) = m ( x - x1)

y = 3.712( x+1/2) add 5 to both sides to get the equation of a tangent line

confidence rating #$&*:3

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Given Solution:

`a Write 25^u where u = 2x^2. So du/dx = 4x.

The derivative of a^x is a^x * ln(a). So the derivative of 25^u with respect to x is

du / dx * ln(25) * 25^u = 4x ln(25) * 25^u = 4x ln(25) * 25^ (2 x^2).

Evaluating this for x = -1/2 you get

4 * (-1/2) ln(25) * 25^(2 * (-1/2)^2 ) = -2 ln(25) * 25^(1/2) = -2 ln(25) * 5 = -10 ln(25) = -20 ln(5) = -32.189 approx.

So the tangent line is a straight line thru (-1/2, 5) and having slope -20 ln(5). The equation of a straight line with slope m passing thru (x1, y1) is

(y - y1) = m ( x - x1) so the slope of the tangent line must be

y - 5 =-20 ln(5) ( x - (-1/2) ) or

y - 5 = -20 ln(5) x - 10 ln(5). Solving for y we get

y = -20 ln(5) x - 10 ln(5).

A decimal approximation is

y = -32.189x - 11.095

ALTERNATIVE SOLUTION:

A straight line has form y - y1 = m ( x - x1), where m is the slope of the graph at the point, which is the value of the derivative of the function at the point. So you have to find the derivative of 25^(2x^2) then evaluate it at x = -1/2.

The derivative of a^x is ln(a) * a^x. The derivative of 25^z would therefore be ln(25) * 25^z. The derivative of 25^(2 x^2) would be found by the chain rule with f(z) = 25^z and g(x) = 2 x^2. The result is g ' (x) * f ' (g(x)) = 4 x * ln(25) * 25^(2x^2). Evaluating at x = -1/2 we get -2 ln(25) * 25^(1/2) = -10 ln(25).

Now we use the ponit-slope form of the equation of a straight line to get (y - 5) = -10 ln(25) * (x - (-1/2) ) and simplify. **

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Self-critique (if necessary):I checked my work several times and came up with the same answer each time, but this seems to be the problem {25^-.5} where you got {25^.5} and I still can’t see where I went wrong. I went through and used {25^.5} and I came up with the same work that you had. I also didn’t rewrite the equation properly when I was isolating y

[(4x)ln(25)][25^(2x^2)]

[(4(-1/2)(ln25)][25^(2(-1/2)^2)]

{-2(ln25)}{25^.5}= -32.189

-2(ln25)*5=-10ln25

y - 5 = -20 ln(5) x - 10 ln(5)

y = -20 ln(5) x - 10 ln(5)

y = -32.189x - 11.095

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Self-critique Rating:3

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Question: `q4.5.25 (previously 4.5.59 (was 4.4.59)) dB = 10 log(I/10^-16); find rate of change when I=10^-4

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Your solution:

dB = 10 log(I/10^-16)

log I = ln(I) / ln(10). According to the rules of logarithums (xloga = ln(x) / ln(a))

(1 / I ) / (1 / ln(10) ) get the derivative

1 / (I ln(10) ) multiplied out the rate of change is

confidence rating #$&*:3

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Given Solution:

`a This function is a composite; the inner function is I / 10^-16, which has derivative 1/10^-16 = 10^16. So the derivative is

dB' = dB / dI = 10 ( 10^16 * / ln(10) ) / (I / 10^-16) = 10 / [ ln(10) * I ].

Alternatively, 10 log(I / 10^-16) = 10 (log I - log(10^-16) ) = 10 log I + 160; the derivative comes out the same with no need of the chain rule.

Plugging in I = 10^-4 we get rate = 10 / [ ln(10) * 10^-4 ] = 10^5 / ln(10), which comes out around 40,000 (use your calculator to get the accurate result. **

STUDENT COMMENT

I did not know how to find the derivative once I simplified the problem. After viewing the

solution, I am still confused.

INSTRUCTOR RESPONSE

log I = ln(I) / ln(10).

The derivative of ln(x) with respect to x is 1/x, so the derivative with respect to I of ln(I) is 1 / I.

So the derivative of ln(I) / ln(10) is (1 / I ) * (1 / ln(10) ) = 1 / (I ln(10) ).

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Self-critique (if necessary):N/A

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Self-critique Rating:

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Question: `q4.5.26 (previously 4.5.60 (was 4.4.60)) T = 87.97 + 34.96 ln p + 7.91 `sqrt(p); find rate of change

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Your solution:

T = 87.97 + 34.96 ln p + 7.91 `sqrt(p)

87.97 goes to 0 34.96(1/p)=34.96/p (7.91 sqrt(p))={7.91^(1/2)}/(p^-1/2) derivative of each section of the function

34.96/p+3.995/p^-1/2 is the derivative

confidence rating #$&*:3

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Given Solution:

`a The derivative with respect to p of ln p is 1 / p and the derivative with respect to p of sqrt(p) is 1 / (2 sqrt(p)). The derivative of the constant 89.97 is zero so

dT/dp = 34.96 * 1/p + 7.91 * 1 / (2 sqrt(p)) = 34.96 / p + 3.955 / sqrt(p). **

STUDENT QUESTION

Not understanding why the sqrt does not go away

INSTRUCTOR RESPONSE

The derivative with respect to p of sqrt(p) is 1 / (2 sqrt(p)).

This is a familiar derivative from first-semester calculus, obtained from the power-function rule that tells us that the derivative of x^a is a x^(a - 1). (this is usually stated with exponent p instead of a, but since p is the variable in this problem that form would almost certainly be confusing).

For example the derivative of x^3 is 3 * x^(3-1) = 3 x^2.

The derivative relevant to the current problem is the derivative of sqrt(x), or x^(1/2). The derivative is 1/2 * x^(1/2 - 1) = 1/2 x ^ (-1/2).

x^(-1/2) = 1 / x^(1/2) = 1 / sqrt(x). Therefore our derivative 1/2 x^(-1/2) is 1 / (2 sqrt(x)).

In the current problem the variable is p rather than x. The derivative, with respect to p, of p^(1/2) is 1/2 p^(1/2 - 1) = 1 / (2 sqrt(p) ).

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Self-critique (if necessary): does it matter if you write it as 3.995/p^-1/2 or 3.995/sqrt(p)

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Self-critique Rating:

@& The meanings are identical.*@

Note that the documents listed below are recommended as a summary, in q_a_ form, of the rules and applications of differentiation. You are welcome to submit any or all of these, or any part of any of these documents. They are recommended for anyone who isn't completely sure of the present chapter.

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa11.htm

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa12.htm

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa13.htm

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa14.htm

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa15.htm

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa16.htm

"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

Note that the documents listed below are recommended as a summary, in q_a_ form, of the rules and applications of differentiation. You are welcome to submit any or all of these, or any part of any of these documents. They are recommended for anyone who isn't completely sure of the present chapter.

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa11.htm

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa12.htm

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa13.htm

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa14.htm

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa15.htm

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa16.htm

"

Self-critique (if necessary):

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Self-critique rating:

#*&!

Note that the documents listed below are recommended as a summary, in q_a_ form, of the rules and applications of differentiation. You are welcome to submit any or all of these, or any part of any of these documents. They are recommended for anyone who isn't completely sure of the present chapter.

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa11.htm

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa12.htm

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa13.htm

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa14.htm

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa15.htm

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa16.htm

"

Self-critique (if necessary):

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Self-critique rating:

#*&!#*&!

&#This looks good. See my notes. Let me know if you have any questions. &#