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Homework Help: Find the y-intercept of the tangent line to: -.4/ sqrt(3 + x) at

  1. Jan 9, 2013 #1
    1.Find the y-intercept of the tangent line to: y= -.4/√(3 + x) at [2.5, -.170560573084488]

    2. So I thought the first step would be to find the slope of the tangent line.
    I think we find the slope of the tangent line by taking the derivative.

    So I am going to use the Quotient Rule to take the derivative.

    [(√3 + x)(0) - (-.4)(.5(x + 3)^-1/2) * 1] / √(x + 3)^2

    .2√(x + 3) / (x + 3) <----- This is the slope of the tan line to y= -.4/√(3 + x)

    If we want to know the slope at x = 2.5, then wouldn't we plug in 2.5 for x?

    So, [.2√(3 + 2.5)]/(2.5 + 3) = .085<----- This is the slope

    Now, I thought we could substitute everything into y = mx + b in order to find out what b is.

    -.1705606 = .085(2.5) + b

    b = -.3831

    3. But this is NOT the right answer!
    Please tell me what I am doing wrong!
    Thank you! :)
  2. jcsd
  3. Jan 9, 2013 #2


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    Your derivative is wrong. What is u^(-1/2)/u? You aren't combining the exponents correctly. u^(-1/2)=1/sqrt(u). Not sqrt(u).
  4. Jan 9, 2013 #3


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    For calculating the derivative, instead of using the quotient rule, rewrite the equation:
    y = -0.4/SQRT(x+3) as y = -0.4 * (x+3)^(-1/2)

    Now, you have an eqn. of the form y = a u^n and you can use the power rule to get the derivative.

    The Quotient Rule is useful when you are dealing with the ratio of two functions.
  5. Jan 9, 2013 #4


    Staff: Mentor

    A better choice (because it's simpler and less prone to errors) would be write your equation as y = -.4 * (x + 3)-1/2
    It's a good idea to keep track of what you're doing. For the above, you're working with the derivative, so this line should start with y' = or dy/dx = .

    I see that you are including brackets around your numerator, so you've been listening. Good!

    You have a mistake above that might just be a transcription error. At the beginning you have (√3 + x)(0) ...
    That should be [(√(3 + x)(0)
    You have a mistake above. You should have gotten
    dy/dx = .2(x + 3)-1/2/(x + 3)

  6. Jan 9, 2013 #5


    Staff: Mentor

    To expand on what SteamKing said, you should NEVER use the quotient rule if either the numerator or denominator is a constant.

    It's not incorrect to use it in this case, but it's more complicated, and the chances are greater of making a mistake.

    For example, if f(x) = 2/(x2 + 1), write this as f(x) = 2(x2 + 1)-1 and use the chain rule to get f'(x) = 2(-1)(x2 + 1)-2(2x).
  7. Jan 9, 2013 #6
    When I was working it out on paper, I left out the (-) on the -1/2 exponent!

    Thanks for pointing that out!

    So the Quotient Rule method is:

    [.4(1/2(x + 3)^-1/2)]/(x + 3)

    .2/(x + 3)^1/2 * (x + 3)

    = .2/ (x + 3)^3/2

    If I substitute the x = 2.5, then I get: .0155 as the slope.

    I tried the Product Rule as well to see if I'd get the same thing.

    So, -.4*-1/2(x + 3)^-3/2 + [(x + 3)^-1/2](0)

    = .2(x +3)^-3/2

    When substituting x = 2.5, I get the same slope of .0155

    When I substitute everything into y = mx + b, I get: -.17056 = .0155(2.5) + b
    b = -.209

    ...But this is not the right answer...
    Do you know what could be wrong?
    Is my derivative still wrong?
    Thanks! :)
  8. Jan 9, 2013 #7


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    Now I don't see anything wrong with what you did. It looks correct to me. I get b=(-0.209324).
  9. Jan 10, 2013 #8
    @Dick - Thank you so much for always helping me! :) You are kind.

    I used the extra digits you found, and NOW the computer counts it as correct!
    Thanks! :D
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