Question

Using differentiation, find the equation of the tangent to the curve y=4+ln(x+1) at the point where x=0

Answer 1

dy/dx= 1/(x+1)

Answer 2

when x=0,
dy/dx= 1/1
dy/dx=1

Answer 3

Yes. I got that. But I don't know how to get the equation?

Answer 4

you need to find the coordinates

Answer 5

using x=0, substitute into original y=4+ln(x+1)
then y=4

Answer 6

you have (0, 4) and gradient, m=1

Answer 7

\[y-y _{1}=m(x-x _{1})\]

Answer 8

Oh, Ok. so the answer is y=x+4 :)

Answer 9

yep

Answer 10

Thank you! :)

Answer 11

no problem :)