SOLUTION: Find an equation of the tangent line to the curve y=e^x/(1+2e^x ) at the point whose x-coordinate is 0. Differentiate f(x)=x^x+e^(x-x^3 )-ln⁡〖(ln⁡〖(3x)〗)〗

Algebra ->  Graphs -> SOLUTION: Find an equation of the tangent line to the curve y=e^x/(1+2e^x ) at the point whose x-coordinate is 0. Differentiate f(x)=x^x+e^(x-x^3 )-ln⁡〖(ln⁡〖(3x)〗)〗       Log On


   

Question 1159793: Find an equation of the tangent line to the curve y=e^x/(1+2e^x ) at the point whose x-coordinate is 0.
Differentiate f(x)=x^x+e^(x-x^3 )-ln⁡〖(ln⁡〖(3x)〗)〗
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
y=e%5Ex%2F%281%2B2e%5Ex+%29
if x=0
find tangent point:
y=e%5E0%2F%281%2B2e%5E0+%29
y=1%2F%281%2B2%2A1+%29
y=1%2F3
tangent point is: (0,1%2F3)


we need a slope of the tangent line, so find derivative:

d%2Fdx%29=e%5Ex%2F%281%2B2e%5Ex+%29
apply the Quotient Rule
y'..........this simplifies to
y'%28x%29+=+1%2F%282+%282e%5Ex+%2B+1%29%29+-+1%2F%282+%282e%5Ex+%2B+1%29%5E2%29

use given x=0+to find the slope of the tangent line

y'%280%29+=+1%2F%282+%282e%5E0+%2B+1%29%29+-+1%2F%282+%282e%5E0+%2B+1%29%5E2%29
y'%280%29+=+1%2F%282+%282%2A1%2B+1%29%29+-+1%2F%282+%282%2A1+%2B+1%29%5E2%29
y'%280%29+=+1%2F6+-+1%2F18
y'%280%29+=+1%2F9=> the slope of the tangent line is m=1%2F9

tangent line is:
y=mx%2Bb..............use tangent point (0,1%2F3) and the slope m=1%2F9
1%2F3=%281%2F9%29%2A0%2Bb
b=1%2F3

and tangent line is:
+y=%281%2F9%29x%2B1%2F3