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Math Help - finding the slope of the tangent line

  1. #1
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    finding the slope of the tangent line

    i know that you can use the point slope formula. but what if all they give for is x=1, for y=e^(-x )+ e^(2x-1). do you find the derv of the function and then plug in 1 to X to find y and then use the point slope formula?
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  2. #2
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    Quote Originally Posted by johnnyc418 View Post
    i know that you can use the point slope formula. but what if all they give for is x=1, for y=e^(-x )+ e^(2x-1). do you find the derv of the function and then plug in 1 to X to find y and then use the point slope formula?
    if you'll find/get the derivative of f(x)=e^(-x )+ e^(2x-1) and look for f'(1);that is,for example, f'(1)=n, then n is the slope of the tangent line at x=1.
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  3. #3
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    just to verify, the derivative is y'=-e^-x + 2 correct? so then i would do
    y'=-e^-1 +2?
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    Quote Originally Posted by johnnyc418 View Post
    just to verify, the derivative is y'=-e^-x + 2 correct? so then i would do
    y'=-e^-1 +2?
    If y=e^(-x)+e^(2x-1) then no that's not the derivative. The second term is wrong. It should be (2)e^(2x-1). Do you see why?

    Once you get the right y', plug in x=1 to get the slope of y at x=1. Then use the point-slope formula: y-y_0=m(x-x_0) to find the tangent line.
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    Quote Originally Posted by Jameson View Post
    If y=e^(-x)+e^(2x-1) then no that's not the derivative. The second term is wrong. It should be (2)e^(2x-1). Do you see why?

    Once you get the right y', plug in x=1 to get the slope of y at x=1. Then use the point-slope formula: y-y_0=m(x-x_0) to find the tangent line.

    no not really. i dont understand how youre getting (2)e^(2x-1)
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  6. #6
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    Quote Originally Posted by johnnyc418 View Post
    no not really. i dont understand how youre getting (2)e^(2x-1)
    \frac{d}{dx}e^{f(x)}=e^{f(x)}*f'(x)

    It's the chain rule property used so often. The derivative of (2x-1) is just 2, so that is why there is a 2 in front now. The exponent of e remains the same.
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  7. #7
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    Quote Originally Posted by Jameson View Post
    \frac{d}{dx}e^{f(x)}=e^{f(x)}*f'(x)

    It's the chain rule property used so often. The derivative of (2x-1) is just 2, so that is why there is a 2 in front now. The exponent of e remains the same.
    what about the first part that says e^-x. isnt that equal to -e^-x? what happens to that?
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  8. #8
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    Quote Originally Posted by johnnyc418 View Post
    what about the first part that says e^-x. isnt that equal to -e^-x? what happens to that?
    I was only referencing the error you made, which was in the 2nd term. Sorry if that was confusing. You are correct that the derivative of e^(-x) is -e^(-x)
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  9. #9
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    so would the derivative be y'=-e^-x + (2)e^(2x-1)?
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  10. #10
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    Quote Originally Posted by johnnyc418 View Post
    so would the derivative be y'=-e^-x + (2)e^(2x-1)?
    Yes.
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