tangent line to y = e^x + ln(x + 1) at (0, 1)

[tinad]

find the tangent line to y=e^x + ln(x+1) at (0,1)
e^x=0 or 1 i thought e^0=1
does the ln(x+1) then become e(x+1) giving y= ex+1?

 

[ChaoticLlama]

AGAIN, I question your statement " if e^x = 0 "

Solve for x in that statment please, and tell me what you find.

Please solve for that mathematical statement that you have posted twice yet never issued any explantion for why you need it.

I do not believe that you are learning anything from us here with the types of posts you have been making. Please post all attempts you have made in order to solve the problem.

We are here to critique and correct solutions, not insult or embaress you.

 

[steve555]

Hey, you first have to find the derivative of the equation.
Which is: y'=e^x + 1/(x+1)
Then subsitute x for 0: y'=e^0 + 1/(0+1) >> y'=1+1 >> y'=2
After that, we plugin the slope for equation of a line. Slope is 2.
Equation of line: y-1=2(x-0) >> y=2x+1
y=2x+1 is your final answer.

Hopefully that helps and I am not too late.