find the equation of the tangent line to the curve at the given point.

y=(1+2x)^10 (0,1)

Printable View

- Oct 13th 2008, 07:43 AMthecountequation of tangent line
find the equation of the tangent line to the curve at the given point.

y=(1+2x)^10 (0,1) - Oct 13th 2008, 08:10 AMmaybeline9216
"y=(1+2x)^10" Is this the equation of the curve??

- Oct 13th 2008, 08:15 AMmaybeline9216
First find what is dy/dx.

Note that the gradient of the tangent to the curve at x=0 is equal to the gradient of the curve at the pt. x=0

To find dy/dx, use chain rule.

dy/dx=10{(1+2x)^9} x 2

=20{(1+2x)^9}

Then sub x=0 into dy/dx.

U will get dy/dx=20

Eqn.of tangent to the curve at x=0

=> [(y-1)/x] = 20

Complete the rest on your own...