# Differentiation and 1st Principles Question

• Mar 12th 2007, 08:30 PM
SportfreundeKeaneKent
Differentiation and 1st Principles Question
I'm not to good at these questions where you have to find the equation and I'm stuck with the one where you have to find the answer w/ 1st principles.

http://img98.imageshack.us/img98/3199/810tz1.png

In case you can't see that pic:

http://img98.imageshack.us/img98/3199/810tz1.png
• Mar 12th 2007, 09:31 PM
Jhevon
Quote:

Originally Posted by SportfreundeKeaneKent
I'm not to good at these questions where you have to find the equation and I'm stuck with the one where you have to find the answer w/ 1st principles.

http://img98.imageshack.us/img98/3199/810tz1.png

In case you can't see that pic:

http://img98.imageshack.us/img98/3199/810tz1.png

8. Remember that the derivative, dy/dx gives you the slope of the tangent line for any given value of x. So let's find dy/dx (or y', means the same thing), we will have to do this using implicit differentiation, i hope you know it.

x^3 + y^3 - 3xy = 17
=> 3x^2 + 3y^2 y' - 3y - 3x y' = 0 ....now we group the y' together
=> 3y^2 y' - 3x y' = -3x^2 + 3y
=> y'(3y^2 - 3x) = 3y -3x^2
=> y' = (3y - 3x^2)/(3y^2 - 3x)

now at (2,3), y' = [3(3) - 3(2)^2]/[3(3)^2 - 3(2)] = (-3)/(21) = -1/7

now recall that the equation of a line is of the form y = mx + b, where m is the slope given by the derivative and b is the y-intercept. we will use the point-slope form to find the equation of the tangent line.

y - y1 = m(x - x1)
=> y - 3 = (-1/7)(x - 2)
=> y = (-1/7)x + 2/7 + 3
=> y = (-1/7)x + 23/7 ............equation of the tangent line.

9. Given the curve y = kx^2 + 5 and the eqution of the tangent line 4x + y = 9

First let's see where these lines intersect. we have:

y = -4x + 9
y = kx^2 + 5

=> kx^2 + 5 = -4x + 9
=> kx^2 + 4x - 4 = 0

=> x = [-4 +/- sqrt(16 + 16k)]/2k ............by the quadratic formula
=> x = [-4 +/- -4sqrt(1 + k)]/2k
=> x = [-2 +/- -2sqrt(1 + k)]/k

now y = kx^2 + 5
=> y' = 2kx
since the slope of the given tangent line is -4, we want y' = 2kx = -4
=> 2kx = -4
=> kx = -2
=> x = -2/k

now substitute this value for x in the formula we made above.
x = [-2 +/- -2sqrt(1 + k)]/k
=> -2/k = [-2 +/- -2sqrt(1 + k)]/k
=> -2 = -2 +/- -2sqrt(1 + k)
=> 0 = +/- -2sqrt(1 + k)
=> 0 = +/- sqrt(1 + k)
=> 0 = 1 + k ................squared both sides
=> k = -1

10. f(x) = sqrt(2 - x)

they want us to find the derivative of f(x) using first principles. this means they want us to find it by the limit definition:

using f ' (x) = limit{h-->0} [f(x + h) - f(x)]/h
=> f ' (x) = limit{h-->0} [sqrt(2 - x - h) - sqrt(2 - x)]/h ...now we abandon our mathematical training for a moment and irrationalize the quotient.
=> f ' (x) = limit{h-->0} [sqrt(2 - x - h) + sqrt(2 - x)]/[sqrt(2 - x - h) + sqrt(2 - x)] * [sqrt(2 - x - h) - sqrt(2 - x)]/h
=> f ' (x) = limit{h-->0} [2 - h - x - (2 - x)]/h[sqrt(2 - x - h) + sqrt(2 - x)]
=> f ' (x) =
limit{h-->0} -h/h[sqrt(2 - x - h) + sqrt(2 - x)]
=> f ' (x) =
limit{h-->0} -1/[sqrt(2 - x - h) + sqrt(2 - x)]
finally we can take the limit without worrying about the denominator going to zero.
=>
f ' (x) = -1/[sqrt(2 - x) + sqrt(2 - x)]
=>
f ' (x) = -1/[2*sqrt(2 - x)]
=> f ' (x) = -(1/2)(2 - x)^(-1/2) ..........as expected

if there's anything you don't understand about any of the questions, dont hesitate to ask
• Mar 16th 2007, 08:53 PM
SportfreundeKeaneKent
I don't understand some parts of the second one (#9)

When you go from
x = [-4 +/- sqrt(16 + 16k)]/2k ............by the quadratic formula

to

=> x = [-4 +/- -4sqrt(1 + k)]/2k

How do you take the (-4) out and put it in front of the square root?

Also, when you go from 0 = +/- -2sqrt(1 + k)
to:
=> 0 = +/- sqrt(1 + k)

Do you divide both sides by (-2) to get rid of the -2?
• Mar 16th 2007, 09:03 PM
DistantCube
x = [-4 +/- sqrt(16 + 16k)]/2k
x = [-4 +/- sqrt[16(1+k)] ]/2k
x = [-4 +/- sqrt(16).sqrt(1+k)]/2k

and you know sqrt(16)= +/- 4
taking either works, since we already have a +/- it would just change it to a -/+

x = [-4 +/- -4sqrt(1+k)]/2k

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to your second question, yes, just divide both side by -2.
• Mar 17th 2007, 02:18 PM
SportfreundeKeaneKent
Sorry to bother you again, but I just can't understand the one step in the third question (#10)

How do you go from this:
=> f ' (x) = limit{h-->0} [sqrt(2 - x - h) - sqrt(2 - x)]/h
to this:
f ' (x) = limit{h-->0} [sqrt(2 - x - h) + sqrt(2 - x)]/[sqrt(2 - x - h) + sqrt(2 - x)] * [sqrt(2 - x - h) - sqrt(2 - x)]/h

Edit: Never mind, I see what you did. I wrote it out again on paper without the sqrt or ^ written and saw it. Thanks anyways :-)