# Thread: Differential and finding an equation.

1. ## Differential and finding an equation.

Find an equation of the tangent line to the curve at the given point.
$y = (1+2x)^{10} , (0,1)$

So I use the chain rule for this don't I?

$y' = 10(1+2x)^{9}(2)$

Pretty much stuck from here.
Thanks in advance.

2. Originally Posted by cloudzer0
Find an equation of the tangent line to the curve at the given point.
$y = (1+2x)^{10} , (0,1)$

So I use the chain rule for this don't I?

$y' = 10(1+2x)^{9}(2)$

Pretty much stuck from here.
Thanks in advance.
From here, substitute x=0 into the derivative:

$y^{/}(0)=20$

Plug the slope and the point into the point-slope form of a line:

$y-(1)=(20)(x-(0))=20x+1$.

Hope this helped you out!

3. Originally Posted by cloudzer0
Find an equation of the tangent line to the curve at the given point.
$y = (1+2x)^{10} , (0,1)$

So I use the chain rule for this don't I?

$y' = 10(1+2x)^{9}(2)$

Pretty much stuck from here.
Thanks in advance.
So now we need to the slope.

To find the slope we evaluate the derivative at x=0

$m=y' = 10(1+2(0))^{9}(2)=20$

Now we use the point slope formula

$y-1=20(x-0) \iff y=20x+1$

Yeah!