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1. Use the chain rule to find the 1st derivative of f(x)= (3+x)/(2-x),
all under the square root.
Remember the chain rule:
f(x) = g(h(x)) ==> f'(x) = g'(h(x)) * h'(x)
For this problem call:
g(h) = sqrt(h) = h^1/2
h(x) = (3+x)/(2-x)
then:
f(x) = g(h(x))
then:
[ remember that (j/k)' = (j'k - jk')/k^2 ]
g'(h) = 1/2 * h^(-1/2) = 1/(2*sqrt(h))
h'(x) = [1*(2-x)-(3+x)*(-1)] / (2-x)^2 =
= [2-x+3+x] / (2-x)^2 =
= 5/(2-x)^2
finally:
f'(x) = g'(h(x)) * h'(x) =
= 1/(2*sqrt(h(x))) * 5/(2-x)^2 =
= (5/2) / [sqrt((3+x)/(2-x)) * (2-x)^2] = (you can stop here)
= (5/2) / [sqrt(3+x) * (2-x)^(2-1/2)] =
= (5/2) / [sqrt(3+x) * (2-x)^(3/2)] =
= (5/2) / [sqrt((3+x)*(2-x)) * (2-x)] =
= (5/2) / [(2-x)*sqrt(6-x-x^2)] =
= 5/(4-2x) * 1/sqrt(6-x-x^2)
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2. Find the second derivative of the function g(x)= (x+1) under square root.
Here we must use the chain rule again:
g(x) = h(i(x))
where:
h(i) = sqrt(i) ==> h'(i) = 1/(2*sqrt(i))
i(x) = (x+1) ==> i'(x) = 1
then:
g'(x) = h'(i(x)) * i'(x) =
= 1/(2*sqrt(i(x))) * 1 =
= 1/(2*sqrt(x+1)) =
= 1/2 * (x+1)^(-1/2)
This is the first derivative of g.
For the second derivative, one more time we need to use the chain rule:
g'(x) = j(k(x))
where:
j(k) = 1/2 * k^(-1/2) ==> j'(k) = -1/4 * k^(-3/2)
k(x) = (x+1) ==> k'(x) = 1
then:
g''(x) = j'(k(x)) * k'(x) =
= -1/4 * k(x)^(-3/2) * 1 =
= -1/4*(x+1)^(-3/2)
This is the second derivative of g(x).
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3. Given y^3 + x^2*y^5-x^4= 27 find
A) dy/dx, (the 1st derivative of y with respect to x):
For simplicity we will call y' = dy/dx
We have that:
y^3 + x^2*y^5-x^4 = 27
then:
(y^3 + x^2*y^5-x^4)'=(27)' ==> (y^3)'+(x^2*y^5)'-(x^4)'=0 ==>
==> 3y^2*y' + (2x*y^5 + x^2*5y^4*y') - 4x^3 = 0 ==>
==> y'*(3y^2 + x^2*5y^4) + 2x*y^5 - 4x^3 = 0 ==>
==> dy/dx = y' = (4x^3 - 2x*y^5) / (3y^2 + x^2*5y^4)
B) find the slope of the tangent line to the curve at point (0,3)
The slope of the tangent line to the curve at point (0,3) is the evaluation of
y' = (4x^3 - 2x*y^5) / (3y^2 + x^2*5y^4) for x=0 and y=3
Slope(0,3) = (4*0^3 - 2*0*3^5) / (3*3^2 + 0^2*5*3^4) = 0
For additional reference, with examples, regarding the chain rule and
its applications, including implicit differentation (this problem),
please visit the following pages:
"Karl's Calculus Tutor - 4.4 Derivatives: Chain Rule Applications":
http://www.karlscalculus.org/calc4_4.html
"Implicit Differentiation":
http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/implicitdiffdirectory/ImplicitDiff.html
"Calculus I (Math 2413) - Derivatives - Implicit Differentiation":
http://tutorial.math.lamar.edu/AllBrowsers/2413/ImplicitDIff.asp
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4. You have a piece of square cardboard that has dimensions of 6
inches by 6 inches. Square corners of dimesion (H) are cut from the
sides. The corners are folded up to make a box.
(Tip = Draw the square cardboard and the square corners)
A) what are the demension of (H) that will yeild a box of maximum volume?
Volume of the box = (A)*(H), where A is the area of its base and H
is the Height.
The size of each side of the base is (6-2H), then A = (6-2H)^2; so we have that:
Volume of the box = V = (6-2H)^2 * H =
= (36 - 24H + H^2)*H =
= H^3 - 24H^2 + 36H
Note that 0 =< H =< 6 , and obviusly at the end points of the interval
the volume is zero (no heigh or no base).
We need to find a valid value of H that makes V'(H) = 0 .
So we need to find V' and equal it to zero:
V' = 3H^2 - 48H + 36 = 0
==> H = [48 + sqrt(48^2 - 4*3*36)]/6 = 15.21 (invalid value)
or
H = [48 - sqrt(48^2 - 4*3*36)]/6 = 0.7889 (valid)
This value of H makes the first derivative zero, to know if it this a
maximum, the second derivative of V at this value must be negative:
V''(H) = 6H - 48 ==> V''(0.7889) < 0 ==> at this value of H the Volume
of the box yield its maximum possible volume.
B) what is the maximum volume?
Just evaluate the Volume formula at H=0.7889 .
V(0.7889) = (0.7889)^3 - 24*(0.7889)^2 + 36*(0.7889) =
= 13.955 cubic inches
For references regarding maximums and minimums see:
"The First Derivative: Maxima and Minima - HMC Calculus Tutorial":
http://www.math.hmc.edu/calculus/tutorials/extrema/
"Calculus - Calculating Minimum and Maximum Values Part I":
http://www.1728.com/minmax.htm
"Calculus - Calculating Minimum and Maximum Values - Part II":
http://www.1728.com/minmax2.htm
"Calculus - Calculating Minimum and Maximum Values - Part III":
http://www.1728.com/minmax3.htm
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5) The demand function for a product is given by p(x)= 1000-x, where p
is the unit price when the demand is x. the cost of producing the
product is determined to be C(x)= 3000+ 20x .
A) find the revenue function
Revenue = price per unit * quantity demanded = p*x =
= (1000-x) * x =
= 1000x - x^2
B) find the profit function
Profit = Revenue - Cost =
= (1000x - x^2) - (3000+ 20x) =
= 980x - x^2 - 3000
C) find the price to be charged in order to maximize the profit
First find the profit's first derivative and equal it to zero:
P'(x) = 980 - 2x = 0 ==> x = 490
Use the second derivative test to check if it is a maximum or a minimum:
P''(x) = -2 ==> P''(490) < 0 it is a maximum.
We have found the demand at which the profit if maximized, we want to
know the correspondent price, to do that just plug the value of x
found in the demand function:
p(x) = 1000-x ==> p(490) = 510 , this is the price to be charged in
order to maximize the profit.
D) find the demand when the profit is maximized
We have found this value at point C), it is x=490
-------------------
6) Using analytical methods find the all the Horizontal and Vertical
Asymptotes of the following function:
H(x)= 3x^3-3x^2+5x-10/5x^3-5x^2+6x
Before starting with this problem please visit the following page:
"Rational Functions and Asymptotes" at Richland Community College:
http://www.richland.edu/james/lecture/m116/polynomials/rational.html
Now we must start with definitions:
The line y=k is an horizontal asymptote of a function f if f(x)-->k
when x tends to infinite (positive or negative).
This definition leads to the following rules for rational functions:
1) If the degree (highest power) of the numerator is larger than the
degree of the denominator, then there is no horizontal asymptote.
2) If the degree of the numerator is smaller than the degree of the
denominator, then the horizontal asymptote is at y = 0 (the x?axis).
3) If the degree of the numerator is equal to the degree of the
denominator, then then y=A/B is the horizontal asymptote, where A is
the coefficient of the highest power of the numerator and B is the
coefficient of the highest power of the denominator.
In this case the degree of the numerator is equal to the degree of the
denominator, then y = 3/5 is the Horizontal Asymptote.
DEFINITION 2:
The line x=c is a vertical asymptote of a function f if:
1) If is undefined at x=c
and
2) f(x) tends to positive or negative infinite as x approaches c from
the left (c-) or from the right (c+).
For rational functions the above definition means that equations of
the vertical asymptotes can be found by finding the roots of the
denominator.
In this case the denominator is 5x^3-5x^2+6x, so we must work in finding its roots:
5x^3-5x^2+6x = x*(5x^2-5x+6)
The discriminant of 5x^2-5x+6 is:
D = 25 - 4*5*6 = 25 - 120 = -95 < 0, then 5x^2-5x+6 has no roots.
So the only root of the denominator is x=0, and this is the only
Vertical asymptote.
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I hope that this helps you. If you find something unclear, or (why
not) a calculation error, or if you need further assistance in one
topic of this question, please do not hesitate to request for an
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to your requests.
Best regards,
livioflores-ga |
