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anonymous

  • one year ago

Help Find the equation of the tangent line to the curve y= 1+x/1+e^x at the point (0,1/2)

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  1. FireKat97
    • one year ago
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    do you know how to find the gradient at a specific point on a curve?

  2. anonymous
    • one year ago
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    No

  3. anonymous
    • one year ago
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    Do you know how to take the derivative of a function?

  4. anonymous
    • one year ago
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    Yes

  5. anonymous
    • one year ago
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    \[\frac{ dy }{ dx }=\frac{ \left( 1+e^x \right)*1-\left( 1+x \right)*e^x }{ \left( 1+e^x \right)^2 }\] \[=\frac{ 1+e^x-e^x-xe^x }{ \left( 1+e^x \right)^2 }=\frac{ 1-xe^x }{ \left( 1+e^x \right)^2 }\] put x=0 and find dy/dx then write the eq. of tangent line \[y-\frac{ 1 }{ 2 }=\frac{ dy }{ dx }\left( x-0 \right)\]

  6. anonymous
    • one year ago
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    I got up to that but the (1+ex)^2 confused me. I know the top is 0

  7. anonymous
    • one year ago
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    At least I think? Or it could be 1

  8. anonymous
    • one year ago
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    top is not zero 1-0*e^0=1-0*1=1

  9. anonymous
    • one year ago
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    bottom is (1+e^0)^2=(1+1)^2=2^2=4

  10. anonymous
    • one year ago
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    So the slope is 1/4?

  11. anonymous
    • one year ago
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    correct.

  12. anonymous
    • one year ago
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    Okay thanks so much

  13. anonymous
    • one year ago
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    yw

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