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anonymous
 one year ago
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Find the equation of the tangent line to the curve y= 1+x/1+e^x at the point (0,1/2)
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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FireKat97
 one year ago
Best ResponseYou've already chosen the best response.0do you know how to find the gradient at a specific point on a curve?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do you know how to take the derivative of a function?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ dy }{ dx }=\frac{ \left( 1+e^x \right)*1\left( 1+x \right)*e^x }{ \left( 1+e^x \right)^2 }\] \[=\frac{ 1+e^xe^xxe^x }{ \left( 1+e^x \right)^2 }=\frac{ 1xe^x }{ \left( 1+e^x \right)^2 }\] put x=0 and find dy/dx then write the eq. of tangent line \[y\frac{ 1 }{ 2 }=\frac{ dy }{ dx }\left( x0 \right)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got up to that but the (1+ex)^2 confused me. I know the top is 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0At least I think? Or it could be 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0top is not zero 10*e^0=10*1=1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0bottom is (1+e^0)^2=(1+1)^2=2^2=4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So the slope is 1/4?
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