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Math Help - Derivatives - Please Check my answers

  1. #1
    Member FalconPUNCH!'s Avatar
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    Derivatives - Please Check my answers

    1) Differentiate

    Y = x^2sinxtanx

    f(x) = sinx ; f'(x) = cosx
    g(x) = tanx ; g'(x) = sec^2

    So I get:

    Y' = 2cosxtanx + x^2sinxsec^2

    2) Find an equation of the tangent line to the curve at the given point:


    P(0,1)

    Y = 1/(sinx + cosx)

    Y' = 2sinx/(sinx + cosx)^2
    Last edited by FalconPUNCH!; October 18th 2007 at 07:23 PM.
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    Member FalconPUNCH!'s Avatar
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    anyone
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by FalconPUNCH! View Post
    1) Differentiate

    Y = x^2sinx + tanx

    f(x) = sinx ; f'(x) = cosx
    g(x) = tanx ; g'(x) = sec^2

    So I get:

    Y' = x^2cosxtanx + x^2sinxsec^2
    this is wrong. you need the product rule for the first term. and the derivative of the second term is simply sec^2(x), where did the x^2sin(x) come from that you attached to it?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by FalconPUNCH! View Post


    2) Find an equation of the tangent line to the curve at the given point:


    P(0,1)

    Y = 1/(sinx + cosx)

    Y' = 2sinx/(sinx + cosx)^2
    this is obviously wrong based on the fact that you haven't answered the question. where is the tangent line? it is also wrong because you did not differentiate properly. use the quotient rule...or if it pleases your fancy, change the function to y = (sin(x) + cos(x))^{-1} and use the chain rule
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    Member FalconPUNCH!'s Avatar
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    Quote Originally Posted by Jhevon View Post
    this is obviously wrong based on the fact that you haven't answered the question. where is the tangent line? it is also wrong because you did not differentiate properly. use the quotient rule...or if it pleases your fancy, change the function to y = (sin(x) + cos(x))^{-1} and use the chain rule
    I don't really understand either of the rules.

    Quote Originally Posted by Jhevon View Post
    this is wrong. you need the product rule for the first term. and the derivative of the second term is simply sec^2(x), where did the x^2sin(x) come from that you attached to it?
    Oops there's not supposed to be a + sign there. It's supposed to be multiplication.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by FalconPUNCH! View Post
    I don't really understand either of the rules.


    Oops there's not supposed to be a + sign there. It's supposed to be multiplication.
    Product Rule: \frac d{dx}f(x)g(x) = f'(x)g(x) + f(x)g'(x)

    Quotient Rule: \frac d{dx} \frac {f(x)}{g(x)} = \frac {g(x)f'(x) - f(x)g'(x)}{[g(x)]^2} .............the order of the top matters, so be sure to apply this exactly

    Chain Rule: \frac d{dx}f(g(x)) = f'(g(x))g'(x)

    Can you continue?
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    Member FalconPUNCH!'s Avatar
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    Quote Originally Posted by Jhevon View Post
    Product Rule: \frac d{dx}f(x)g(x) = f'(x)g(x) + f(x)g'(x)

    Quotient Rule: \frac d{dx} \frac {f(x)}{g(x)} = \frac {g(x)f'(x) - f(x)g'(x)}{[g(x)]^2} .............the order of the top matters, so be sure to apply this exactly

    Chain Rule: \frac d{dx}f(g(x)) = f'(g(x))g'(x)

    Can you continue?
    I got this for the first one:

    Y' = 2cosxtanx + x^2sinxsec^2

    and this for the second one:

    Y' = cosx - sinx/(sinx + cosx)^2
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by FalconPUNCH! View Post
    1) Differentiate

    Y = x^2sinxtanx

    f(x) = sinx ; f'(x) = cosx
    g(x) = tanx ; g'(x) = sec^2

    So I get:

    Y' = 2cosxtanx + x^2sinxsec^2

    we have three product here, you would have to take two at a time, so by the product rule you would say:

    y' = \left( x^2 \sin x \right)' \tan x + x^2 \sin x (\tan x )'

    but for \left( x^2 \sin x \right)' you would use the product rule again, to get, in total:

    y' = \left( 2x \sin x + x^2 \cos x \right) \tan x + x^2 \sin x ( \tan x )'


    ALTERNATIVELY:

    you could have realized that:

    y = \frac {x^2 \sin^2 x}{\cos x}

    and use the quotient rule




    2) Find an equation of the tangent line to the curve at the given point:


    P(0,1)

    Y = 1/(sinx + cosx)

    Y' = 2sinx/(sinx + cosx)^2
    you answer to the second was correct, but your answer is still incomplete. you were asked for the tangent line at a point
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    Member FalconPUNCH!'s Avatar
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    Quote Originally Posted by Jhevon View Post
    we have three product here, you would have to take two at a time, so by the product rule you would say:



    you answer to the second was correct, but your answer is still incomplete. you were asked for the tangent line at a point
    I wasn't quite sure about it being three products so I got confused. Yeah I don't know how to find the tangent line at the point. I've checked my notes and the book and it doesn't say anything about it.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by FalconPUNCH! View Post
    I wasn't quite sure about it being three products so I got confused. Yeah I don't know how to find the tangent line at the point. I've checked my notes and the book and it doesn't say anything about it.
    the derivative gives the formula formula for the slope at any value of x. plug in the required x-value into your derivative equation and that will give you the slope at that point. this will be the m in the equation of your line, y = mx + b

    once you have m, you can use the point-slope form:

    y - y_1 = m(x - x_1)

    where (x_1,y_1) is a point the line passes through, this is given to you. just solve for y and you have your line
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  11. #11
    Member FalconPUNCH!'s Avatar
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    Quote Originally Posted by Jhevon View Post
    the derivative gives the formula formula for the slope at any value of x. plug in the required x-value into your derivative equation and that will give you the slope at that point. this will be the m in the equation of your line, y = mx + b

    once you have m, you can use the point-slope form:

    y - y_1 = m(x - x_1)

    where (x_1,y_1) is a point the line passes through, this is given to you. just solve for y and you have your line
    I plugged in 0 and got m = 1. And my equation came out to 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 FalconPUNCH! View Post
    I plugged in 0 and got m = 1. And my equation came out to be y = (x-1)
    that is incorrect, you should have: y - 1 = (x - 0) \implies y = x + 1
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  13. #13
    Member FalconPUNCH!'s Avatar
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    Quote Originally Posted by Jhevon View Post
    that is incorrect, you should have: y - 1 = (x - 0) \implies y = x + 1
    hmmm...oh I see. Thank you a lot. I'm going to do it all over again to make sure I understand.
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