Question

Ok… Tangent line equations are killing me. How do I find the line tangent to y=(x+3)^1/3 at x=-3 ? Thanks

Answer 1

not as hard as it looks
find the derivative, then evaluate it at \(x=-3\) to find your slope

Answer 2

since you already have it written in exponential form, the derivative is easy to find
\[y=(x+3)^{\frac{1}{3}}\]
\[y'=\frac{1}{3}(x+3)^{-\frac{2}{3}}\] but to evaluate this you should put it in radical form as
\[y'=\frac{1}{3\sqrt[3]{(x+3)^2}}\]