Question

Help

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

Answer 1

do you know how to find the gradient at a specific point on a curve?

Answer 2

No

Answer 3

Do you know how to take the derivative of a function?

Answer 4

Yes

Answer 5

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

Answer 6

I got up to that but the (1+ex)^2 confused me. I know the top is 0

Answer 7

At least I think? Or it could be 1

Answer 8

top is not zero
1-0*e^0=1-0*1=1

Answer 9

bottom is (1+e^0)^2=(1+1)^2=2^2=4

Answer 10

So the slope is 1/4?

Answer 11

correct.

Answer 12

Okay thanks so much

Answer 13

yw