- #1
Chen
- 977
- 1
y(x) is defined by this equation:
[tex]\int_0^y{\sqrt[3]{1+t^2} dt} + \int_x^0{(2t+1)^2 dt} = 0[/tex]
How do I find the tangent line to y(x) at x=0?
Here's what I tried to do:
[tex]y(x) = \int_0^y{\sqrt[3]{1+t^2} dt} = \int_0^x{(2t+1)^2 dt}[/tex]
Taking the derivative of both sides of the equation:
[tex]y'(x) = ... = (2x+1)^2 = 4x^2 + 4x + 1[/tex]
[tex]y'(x=0) = 1[/tex]
First, is this method correct?
Second, how do I find y(x=0)? I'm guessing that it's 0, but I'm not sure.
Thanks,
Chen
[tex]\int_0^y{\sqrt[3]{1+t^2} dt} + \int_x^0{(2t+1)^2 dt} = 0[/tex]
How do I find the tangent line to y(x) at x=0?
Here's what I tried to do:
[tex]y(x) = \int_0^y{\sqrt[3]{1+t^2} dt} = \int_0^x{(2t+1)^2 dt}[/tex]
Taking the derivative of both sides of the equation:
[tex]y'(x) = ... = (2x+1)^2 = 4x^2 + 4x + 1[/tex]
[tex]y'(x=0) = 1[/tex]
First, is this method correct?
Second, how do I find y(x=0)? I'm guessing that it's 0, but I'm not sure.
Thanks,
Chen