# 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.

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What is the question? – Riccardo.Alestra Jul 4 '12 at 14:18

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 $$y=4(x_1+1)x+2(1-x_1^2)$$

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But the question as phrased above was not "Find the equation of the tangent line", but rather "Find the slope of the tangent line." – Michael Hardy Jul 4 '12 at 16:37
This answer is wrong. It correctly finds the equation of the tangent line, but that was not the question. The question was how to use the definition of derivative to find the slope (the slope, not the equation) of the tangent line. This answer asserts that the limit as $h\to0$ is $4(x_1+1)$, but the question is how to show that that is the limit. – Michael Hardy Jul 4 '12 at 17:00
Thanks for your answer, you are right. – Vinicius L. Beserra Jul 4 '12 at 19:26
@ViniciusL.Beserra : It's not the same as yours, because yours is not an equation. There's no "$=$" in yours. – Michael Hardy Jul 4 '12 at 22:10

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 $$2\lim_{h\to0}\frac{h(2x_1+h+2)}{h} = 2\lim_{h\to0}(2x_1+h+2) = 2(2x_1+2) = 4x_1+4.$$

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In order to facilitate the calculus I first developed (x+1)^2 first finding. 2x^2+2x+2, and after 2(2x^2+2x+2), I used the denifition to find the answer. – Vinicius L. Beserra Jul 4 '12 at 19:53

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?

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The function was given as $f(x)=2(x+1)^2$, so the derivative is $f'(x)=4x+4$. Besides, the question was how to use the definition of derivative to find this. You haven't done that nor even attempted it. – Michael Hardy Jul 4 '12 at 16:39
The function was written differently before your edit $(2(x-1)^2)$, and I wrote exactly what was necessary to deduce the correct answer given what little he had written. – Eugene Shvarts Jul 4 '12 at 19:58
Yes, i have done and i just reached this same answer of your altough not put it in the question. My doubt is how to find the equation of the slope. So, now for it´s clear that I have to use y1 in the original equation 2(x+1)^2 to find the y1. – Vinicius L. Beserra Jul 4 '12 at 20:05