Question

How do you find the equation of the tangent line to the curve `y=(1+2x)^10` at (0,1)?

Answer 1

Please the explanation of how you do it.

Explanation:

The slope, m. of the tangent line is:

`m = y'(x_0)`

where `x_0` is the x coordinate of the given point.

Use the chain rule to compute y'(x)

`y'(x) = y'(u(x)) = ((dy)/(du))((du)/(dx))`

let `u = 1 + 2x`, then `(du)/(dx) = 2, y(u) = u^10 and (dy)/(du) = 10u^9`

Substituting into the chain rule:

`y'(x) = (10u^9)(2)`

Reverse the u substitution:

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

Evaluate at `x_0 = 0`

`m = y'(0) = 20(1)^9`

`m = 20`

Because the given point is also the y intercept, we may use the slope-intercept form, `y = mx + b`, and we know that b is 1:

`y = 20x + 1`

Answer 2

The equation is `y = 20x + 1`.

Explanation:

Instead of expanding, and getting a massive eleven term expansion, we can simply differentiate the function using the chain rule.

Letting `y = u^10` and `u = 1 + 2x`, `dy/dx = 10u^9 xx 2 = 10(1 + 2x)^9 xx 2 = 20(1 + 2x)^9`

We can now determine the slope of the tangent by substituting the point `x = a` into the derivative.

`m_"tangent" = 20(1 + 0)^9 = 20`

Now, we have the information necessary to find the equation.

`y - y_1 = m(x - x_1)`

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

`y = 20x + 1`

Hopefully this helps!