## 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)

1. FireKat97

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

2. anonymous

No

3. anonymous

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

4. anonymous

Yes

5. 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 $y-\frac{ 1 }{ 2 }=\frac{ dy }{ dx }\left( x-0 \right)$

6. anonymous

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

7. anonymous

At least I think? Or it could be 1

8. anonymous

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

9. anonymous

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

10. anonymous

So the slope is 1/4?

11. anonymous

correct.

12. anonymous

Okay thanks so much

13. anonymous

yw