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
Question: `q 4.5.5 (previously 4.5.10 (was 4.4.10)) find the derivative of ln(1-x)^(1/3)
Your solution:
Confidence Assessment:
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) ].**
Self-critique (if necessary):
Self-critique Rating:
Question: `q4.5.9 (previously 4.5.25 (was 4.4.24)) find the derivative of ln( (e^x + e^-x) / 2)
Your solution:
Confidence Assessment:
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). **
Self-critique (if necessary):
Self-critique Rating:
Question: `q 4.5.10 (previously 4.5.30 (was 4.4.30) ) write log{base 3}(x) with base e
Your solution:
Confidence Assessment:
Given Solution:
`a
We know that
log{base b}(a) =
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.
Self-critique (if necessary):
Self-critique Rating:
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)
Your solution:
Confidence Assessment:
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. **
Self-critique (if necessary):
Self-critique Rating:
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
Your solution:
Confidence Assessment:
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)
).
Self-critique (if necessary):
Self-critique Rating:
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
Your solution:
Confidence Assessment:
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) ).
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