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

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Answer 1 · 2016-10-28T18:34:02

Please the explanation of how you do it.

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 · 2016-10-28T18:41:10

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

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!