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Math Help - More Derivatives

  1. #1
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    More Derivatives

    f(x) = ln(4x^2 +5x-3) - I think you find the deriv of the outside, put the inside back..then find the deriv of the inside?

    So my soln would look like: 1/x(4x^2+5x+3) * (8x+5)? is that right?

    6xsecx+xtan(x) - so we use the product rule for 6xsecx..then product rule for xtanx also?

    (6x)(secx) = (6)(secx)+(6x)(secxtanx) + (1)(tanx)+(x)(sec^2x)?

    And I wrote in my previous post a problem that involved a diver diving off a platfrom 16ft with initial velocity of 32ft/sec...and someone told me it couldn't be done because u need to know the direction of the initial velocity..But I received the same problem on the test and yea..couldn't do it..but why can my teacher do a problem that is said to be undoable?

    She mentioned something about usiing 9.8ft/sec or something and b/2a...so can anyone also explain this to me?

    Thanks
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  2. #2
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    Quote Originally Posted by ktpinnock View Post
    f(x) = ln(4x^2 +5x-3) - I think you find the deriv of the outside, put the inside back..then find the deriv of the inside?

    So my soln would look like: 1/x(4x^2+5x+3) * (8x+5)? is that right?

    6xsecx+xtan(x) - so we use the product rule for 6xsecx..then product rule for xtanx also?

    (6x)(secx) = (6)(secx)+(6x)(secxtanx) + (1)(tanx)+(x)(sec^2x)?

    And I wrote in my previous post a problem that involved a diver diving off a platfrom 16ft with initial velocity of 32ft/sec...and someone told me it couldn't be done because u need to know the direction of the initial velocity..But I received the same problem on the test and yea..couldn't do it..but why can my teacher do a problem that is said to be undoable?

    She mentioned something about usiing 9.8ft/sec or something and b/2a...so can anyone also explain this to me?

    Thanks
    for ln functions in general

    g(x) = \ln f(x)

    g'(x) = \frac{f'(x)}{f(x)}

    so

    f(x) = \ln ( 4x^2+5x-3)

    f'(x) = \frac{\frac{d}{dx}(4x^2+5x-3)}{4x^2 + 5x -3 }

    f'(x) = \frac{8x +5 }{4x^2 +5x-3}

    you derived the second correctly
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  3. #3
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    Quote Originally Posted by ktpinnock View Post
    f(x) = ln(4x^2 +5x-3) - I think you find the deriv of the outside, put the inside back..then find the deriv of the inside?
    Yes, that's the chain rule \frac{df(u(x))}{dx}= \frac{df(u)}{du}\frac{du}{dx}.

    So my soln would look like: 1/x(4x^2+5x+3) * (8x+5)? is that right?
    No. The derivative of ln(u) with respect to u is 1/u so with u= 4x^2+ 5x+ 3 that is \frac{1}{4x^2+ 5x+ 3}. You have an extra "x" in the denominator. You replace the denominator by u, not multiply it.

    6xsecx+xtan(x) - so we use the product rule for 6xsecx..then product rule for xtanx also?
    Yes.

    (6x)(secx) = (6)(secx)+(6x)(secxtanx) + (1)(tanx)+(x)(sec^2x)?
    Yes! very good.

    And I wrote in my previous post a problem that involved a diver diving off a platfrom 16ft with initial velocity of 32ft/sec...and someone told me it couldn't be done because u need to know the direction of the initial velocity..But I received the same problem on the test and yea..couldn't do it..but why can my teacher do a problem that is said to be undoable?

    She mentioned something about usiing 9.8ft/sec or something and b/2a...so can anyone also explain this to me?

    Thanks
    Not unless you explain what she did! I presume she did use a specific direction for the initial motion. Either she did not mention it or you didn't notice. It has been my experience that one begins a dive by going up. Actually, if you don't want to hit the platform on the way down (remember when that happened to Greg Luganis, one of the greatest divers ever, in the Olympics?), you had better angle your initial motion slightly out from the board!

    But since you mention no angles, I suspect your teacher took the easy case of jumping directly up (and then the platform magically disappears!). Your mention of "b/2a" reminds me of completing the square. The equation for height in this situation is a quadratic- its graph is a parabola- and you can find the highest point, the vertex of the parabola, by completing the square.
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    Yes, that's the chain rule \frac{df(u(x))}{dx}= \frac{df(u)}{du}\frac{du}{dx}.


    No. The derivative of ln(u) with respect to u is 1/u so with u= 4x^2+ 5x+ 3 that is \frac{1}{4x^2+ 5x+ 3}. You have an extra "x" in the denominator. You replace the denominator by u, not multiply it.


    Yes.


    Yes! very good.


    Not unless you explain what she did! I presume she did use a specific direction for the initial motion. Either she did not mention it or you didn't notice. It has been my experience that one begins a dive by going up. Actually, if you don't want to hit the platform on the way down (remember when that happened to Greg Luganis, one of the greatest divers ever, in the Olympics?), you had better angle your initial motion slightly out from the board!

    But since you mention no angles, I suspect your teacher took the easy case of jumping directly up (and then the platform magically disappears!). Your mention of "b/2a" reminds me of completing the square. The equation for height in this situation is a quadratic- its graph is a parabola- and you can find the highest point, the vertex of the parabola, by completing the square.
    Alrigt let me just give you the problem I posted yesterday.
    It's straight from her review sheet so I didn't miss anything.

    A dive jumps from a platform diving board that is 32ft above the water with initial velocity of 16ft/sec.
    a. When does the driver hit water? and other questions about velocity, speed, and acceleration. I know for velocity you do the derivative of the function, and speed is the absolute value of velocity, and acceleration is the 2nd derivative of the derivative. But I don't know how to find the function...she said something about an equation -16t^2...and finding time...lol idk that part was somewhat blurry..
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