anonymous
  • anonymous
Help Find the equation of the tangent line to the curve y= 1+x/1+e^x at the point (0,1/2)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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FireKat97
  • FireKat97
do you know how to find the gradient at a specific point on a curve?
anonymous
  • anonymous
No
anonymous
  • anonymous
Do you know how to take the derivative of a function?

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anonymous
  • anonymous
Yes
anonymous
  • anonymous
\[\frac{ dy }{ dx }=\frac{ \left( 1+e^x \right)*1-\left( 1+x \right)*e^x }{ \left( 1+e^x \right)^2 }\] \[=\frac{ 1+e^x-e^x-xe^x }{ \left( 1+e^x \right)^2 }=\frac{ 1-xe^x }{ \left( 1+e^x \right)^2 }\] put x=0 and find dy/dx then write the eq. of tangent line \[y-\frac{ 1 }{ 2 }=\frac{ dy }{ dx }\left( x-0 \right)\]
anonymous
  • anonymous
I got up to that but the (1+ex)^2 confused me. I know the top is 0
anonymous
  • anonymous
At least I think? Or it could be 1
anonymous
  • anonymous
top is not zero 1-0*e^0=1-0*1=1
anonymous
  • anonymous
bottom is (1+e^0)^2=(1+1)^2=2^2=4
anonymous
  • anonymous
So the slope is 1/4?
anonymous
  • anonymous
correct.
anonymous
  • anonymous
Okay thanks so much
anonymous
  • anonymous
yw

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