Trying to solve a slope tangent queston. Given the equation $f(x)=2 (x+1)^2$, use the definition $\dfrac{df}{dx}=\displaystyle\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$ of derivative to find the slope of the tangent to the graph $f(x)$ at point $(x_1, y_1)$; I have reached the following answer,
$$y-2x_1^2-2+4x-4x_1$$
I really, really tried to solve but I´m not certain about the if the answer is right.
 A: Use the point-slope formula.
However, to get the slope using the limit formula we have
$f'(x_1)=\lim_{h\rightarrow 0}\frac{4x_1h+4h+2h^2}{h}=4(x_1+1)$.
So the formula for the tangent line is $y-y_1=4(x_1+1)(x-x_1)$. 
Since $y_1=2(x_1+1)^2$ then the tangent line equation  simplifies to
$$y=4(x_1+1)x+2(1-x_1^2)$$ 
A: If $f(x)=2(x+1)^2$ then
$$
\lim_{h\to0}\frac{f(x_1+h)-f(x_1)}{h} = \lim_{h\to0}\frac{2(x_1+h+1)^2-2(x_1+1)^2}{h}.
$$
The $2$ is a constant and may therefore be pulled out of the limit.  (And remember: in this context, "constant"  means not depending on $h$.)  A bit of trivial algebra---expanding the two squares---says this is equal to
$$
2\lim_{h\to0}\frac{(x_1^2+2x_1h+2x_1+h^2+2h+1)-(x_1^2+2x_1+1)}{h}
$$
Cancelations in the numerator reduce this to
$$
2\lim_{h\to0}\frac{2x_1h+h^2+2h}{h}.
$$
Then this becomes
$$
2\lim_{h\to0}\frac{h(2x_1+h+2)}{h} = 2\lim_{h\to0}(2x_1+h+2) = 2(2x_1+2) = 4x_1+4.
$$
A: Point-slope form for the tangent line will immediately yield $y - y_1 = f'(x_1) (x - x_1)$ as the tangent line. If you've found the derivative correctly, $f'(x) = 4x - 4$, and so $f'(x_1) = 4x_1-4$.
Think you can fix any errors from there?
