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Math Help - Beginning Derivatives.

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    Beginning Derivatives.

    Find an equation for the tangent to the curve at the given point.

    y=2(sqrt(x)) point- (1,2)

    Ok so I think I can work most of it then I get stuck. I am not sure how you guys use the codes so I will try to type it out the best I can.

    lim = ( f(a+h)-f(a) ) / (h)
    h->0
    = f(1+h)-f(1) / h
    = (2 sqrt(1+h) - 2) / h

    This is where I have gotten to. substituting in now will give me a 0 in my denominator. So where would I go next?
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    Quote Originally Posted by ur5pointos2slo View Post
    Find an equation for the tangent to the curve at the given point.

    y=2(sqrt(x)) point- (1,2)

    Ok so I think I can work most of it then I get stuck. I am not sure how you guys use the codes so I will try to type it out the best I can.
    see the first post here.


    lim = ( f(a+h)-f(a) ) / (h)
    h->0
    = f(1+h)-f(1) / h
    = (2 sqrt(1+h) - 2) / h

    This is where I have gotten to. substituting in now will give me a 0 in my denominator. So where would I go next?
    good. now multiply by the conjugate of the numerator over itself. this is a standard trick. (note that the top becomes the factorized for of the difference of two squares).

    so you have \lim_{h \to 0} \frac {2 \sqrt{1 + h} - 2}h \cdot \frac {2 \sqrt{1 + h} + 2}{2 \sqrt{1 + h} + 2}

    now simplify and continue
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    Quote Originally Posted by Jhevon View Post
    see the first post here.

    good. now multiply by the conjugate of the numerator over itself. this is a standard trick. (note that the top becomes the factorized for of the difference of two squares).

    so you have \lim_{h \to 0} \frac {2 \sqrt{1 + h} - 2}h \cdot \frac {2 \sqrt{1 + h} + 2}{2 \sqrt{1 + h} + 2}

    now simplify and continue
    Thanks for that Ill definately take a look at that posting.

    Ok I am just starting but I do remember something about the conjugate from algebra. Would I get..

    ( 4(1+h)-4) / h(2 sqrt(1+h) + 2) ?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ur5pointos2slo View Post
    Thanks for that Ill definately take a look at that posting.

    Ok I am just starting but I do remember something about the conjugate from algebra. Would I get..

    ( 4(1+h)-4) / h(2 sqrt(1+h) + 2) ?
    yes
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    ( 4(1+h)-4) / h(2 sqrt(1+h) + 2)

    Yes well if i plug in for h now im still going to end up with 0 in my denominator..hmmm I am totally lost
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ur5pointos2slo View Post
    ( 4(1+h)-4) / h(2 sqrt(1+h) + 2)

    Yes well if i plug in for h now im still going to end up with 0 in my denominator..hmmm I am totally lost
    i told you to "simplify and continue"

    surely you can simplify the numerator more. you problem h will cancel. just simplify
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    Quote Originally Posted by Jhevon View Post
    i told you to "simplify and continue"

    surely you can simplify the numerator more. you problem h will cancel. just simplify
    ok let me see..

    so 4 (1+h) - 4 /(h(2 sqrt(1+h) + 2) will become

    4h/ ( h ( 2 sqrt(1+h) + 2) ) so can I distribute the h through on the bottom to get...

    4h / 2h + h sqrt(1+h) + 2h? or would this be the wrong direction to go?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ur5pointos2slo View Post
    ok let me see..

    so 4 (1+h) - 4 /(h(2 sqrt(1+h) + 2) will become

    4h/ ( h ( 2 sqrt(1+h) + 2) ) so can I distribute the h through on the bottom to get...

    4h / 2h + h sqrt(1+h) + 2h? or would this be the wrong direction to go?
    NO! why distribute the h??

    you have 4h in the top and an h in the bottom, cancel the h's.
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    Quote Originally Posted by Jhevon View Post
    NO! why distribute the h??

    you have 4h in the top and an h in the bottom, cancel the h's.
    stupid me.. so ill end up with 4/4 which = 1

    which is the slope. so my equation will be y= x+1
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ur5pointos2slo View Post
    stupid me.. so ill end up with 4/4 which = 1

    which is the slope. so my equation will be y= x+1
    yup
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