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Math Help - Chain Rule Derivative (Finding the equation for the tangent line)

  1. #1
    Member FalconPUNCH!'s Avatar
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    Chain Rule Derivative (Finding the equation for the tangent line)

    The equation is  y = x^2 e^{-x} with the point (1, 1/e)

    I used the chain rule (I hope I used it correctly)

     f(x) = x^2 , f'(x) = 2x
     g(x) = e^{-x} ,  g'(x) = e^{-x}

    y' = 2(e^{-x}) e^{-x}

    The problem is I don't know how to find the equation for the tangent line to the curve at the given point.
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  2. #2
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    That's the product rule, not the chain rule.

    \frac{d}{dx}[x^{2}e^{-x}]=(2x-x^{2})e^{-x}

    If x=1, then the slope at that point is \frac{1}{e}

    Now, use your given points in y=mx+b to find the line equation.
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  3. #3
    Member FalconPUNCH!'s Avatar
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    Quote Originally Posted by galactus View Post
    That's the product rule, not the chain rule.

    \frac{d}{dx}[x^{2}e^{-x}]=(2x-x^{2})e^{-x}

    If x=1, then the slope at that point is \frac{1}{e}

    Now, use your given points in y=mx+b to find the line equation.
    I don't understand how you got that equation. Can you explain it to me? Sorry if it's asking too much.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by galactus View Post
    That's the product rule, not the chain rule.

    \frac{d}{dx}[x^{2}e^{-x}]=(2x-x^{2})e^{-x}

    If x=1, then the slope at that point is \frac{1}{e}

    Now, use your given points in y=mx+b to find the line equation.
    Quote Originally Posted by FalconPUNCH! View Post
    I don't understand how you got that equation. Can you explain it to me? Sorry if it's asking too much.
    Okay, so you have the slope of the tangent line at the point where x = 1. What is the y value of this point?

    y = 1^2e^{-1} = \frac{1}{e}

    Thus the tangent line at x = 1 has a slope of \frac{1}{e} and touches the curve at the point \left ( 1, \frac{1}{e} \right ). (There are a couple of 1/e's floating around. Make sure you keep track of them.)

    So your line equation, y = mx + b becomes
    y = \frac{1}{e} \cdot x + b
    at the point \left ( 1, \frac{1}{e} \right ).

    Plug this x and value into the line equation and solve for b.

    -Dan
    Attached Thumbnails Attached Thumbnails Chain Rule Derivative (Finding the equation for the tangent line)-exponential.jpg  
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  5. #5
    Member FalconPUNCH!'s Avatar
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    Quote Originally Posted by topsquark View Post
    Okay, so you have the slope of the tangent line at the point where x = 1. What is the y value of this point?

    y = 1^2e^{-1} = \frac{1}{e}

    Thus the tangent line at x = 1 has a slope of \frac{1}{e} and touches the curve at the point \left ( 1, \frac{1}{e} \right ). (There are a couple of 1/e's floating around. Make sure you keep track of them.)

    So your line equation, y = mx + b becomes
    y = \frac{1}{e} \cdot x + b
    at the point \left ( 1, \frac{1}{e} \right ).

    Plug this x and value into the line equation and solve for b.

    -Dan
    I understand that but I don't know how he got the derivative using the chain rule. I used the chain rule and got the answer in my original post. I don't know what I did wrong.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by FalconPUNCH! View Post
    I understand that but I don't know how he got the derivative using the chain rule. I used the chain rule and got the answer in my original post. I don't know what I did wrong.
    Quote Originally Posted by galactus View Post
    That's the product rule, not the chain rule.

    \frac{d}{dx}[x^{2}e^{-x}]=(2x-x^{2})e^{-x}

    If x=1, then the slope at that point is \frac{1}{e}

    Now, use your given points in y=mx+b to find the line equation.
    The product rule says if we have a function f(x)g(x) then the derivative of that function is
    f^{\prime}(x)g(x) + f(x) g^{\prime}(x)

    Here we have x^2 e^{-x}, so
    f(x) = x^2 \implies f^{\prime}(x) = 2x
    and
    g(x) = e^{-x} \implies g^{\prime}(x) = -e^{-x}

    So the derivative of x^2 e^{-x} will be:
    (2x)(e^{-x}) + (x^2)(-e^{-x}) = (2x - x^2)e^{-x}

    -Dan
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  7. #7
    Member FalconPUNCH!'s Avatar
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    Quote Originally Posted by topsquark View Post
    The product rule says if we have a function f(x)g(x) then the derivative of that function is
    f^{\prime}(x)g(x) + f(x) g^{\prime}(x)

    Here we have x^2 e^{-x}, so
    f(x) = x^2 \implies f^{\prime}(x) = 2x
    and
    g(x) = e^{-x} \implies g^{\prime}(x) = -e^{-x}

    So the derivative of x^2 e^{-x} will be:
    (2x)(e^{-x}) + (x^2)(-e^{-x}) = (2x - x^2)e^{-x}

    -Dan
    Thank you for your help. I'll talk to my teacher because I'm not really sure what the book is asking. The section is about the chain rule so I assumed that you had to use it.
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by FalconPUNCH! View Post
    I understand that but I don't know how he got the derivative using the chain rule. I used the chain rule and got the answer in my original post. I don't know what I did wrong.
    Quote Originally Posted by galactus View Post
    That's the product rule, not the chain rule.

    \frac{d}{dx}[x^{2}e^{-x}]=(2x-x^{2})e^{-x}

    If x=1, then the slope at that point is \frac{1}{e}

    Now, use your given points in y=mx+b to find the line equation.
    You sort of do when you take the derivative of the exponential, but that's only when you don't block it out fully, like I did.

    -Dan

    Quote Originally Posted by FalconPUNCH! View Post
    Thank you for your help. I'll talk to my teacher because I'm not really sure what the book is asking. The section is about the chain rule so I assumed that you had to use it.
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