MHB How Does the Derivative of a Cubic Function Result in a Tangent Line?

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The discussion clarifies that the derivative of a function provides the slope of the tangent line at a specific point, rather than the tangent line itself. For the cubic function y=x^3, the derivative y'=3x^2 yields the slope at any point on the curve. The tangent line can then be constructed using the point-slope formula, incorporating the slope derived from the derivative. This understanding aligns with the general linear equation y=mx+c, where the derivative gives the necessary slope (m) for the tangent line. Overall, the key takeaway is that the derivative indicates the slope at a point, enabling the construction of the tangent line.
tenman
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I'm have trouble understanding a fundamental question of a derivative. So a derivate gives me a tangent line at any given point on a function.

this makes sense for me for a function y=x^2 because the derivative is y'=2x which is a straight line function.

But what about y=x^3 where the derivative is y'=3x^2 this function is a parabola how can a parabola give me a tangent line?

I'm having trouble understand this very basic concept
 
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The first derivative of a function gives you the slope of the tangent line, not the tangent line itself. Thus,for some function $f(x)$, the tangent line (using the point-slope formula) at the point $(x_0,f(x_0))$ is given by:

$$y=f'(x_0)(x-x_0)+f(x_0)$$
 
Ahh ok thanks I think I get it now.

So the general formula for a straight line is y=mx+c

are you saying that the derivative of a function gives you the m value needed to plug into the straight line formula?
 
tenman said:
Ahh ok thanks I think I get it now.

So the general formula for a straight line is y=mx+c

are you saying that the derivative of a function gives you the m value needed to plug into the straight line formula?

Yes, the derivative of the function evaluated at the $x$-coordinate of the tangent point will give you the slope $m$ of the tangent line. :)
 
Thanks heaps for that!