Question

Question #2577a

Answer 1

Let `(x_1,y_1)=(0,1)`.

`y=(1+2x)^{10}`

By Chain Rule,

`y'=10(1+2x)^9cdot(2)=20(1+2x)^9`

So, the slope `m` of the tangent line can be found by

`m=y'(0)=20[1+2(0)]^9=20`

By Point-Slope Form `y-y_1=m(x-x_1)`,

`y-1=20(x-0)`

by adding `1`,

`Rightarrow y=20x+1`

I hope that this was helpful.