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

2 Answers
Oct 28, 2016

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

Oct 28, 2016

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!