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

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SUMMARY

The discussion centers on understanding how the derivative of a cubic function, specifically y=x^3, results in a tangent line. The derivative, y'=3x^2, represents the slope of the tangent line at any given point on the curve. The tangent line can be expressed using the point-slope formula, which incorporates the derivative to determine the slope (m) at a specific point (x_0, f(x_0)). This clarifies that while the derivative itself is not the tangent line, it provides the necessary slope to construct the tangent line equation.

PREREQUISITES
  • Understanding of basic calculus concepts, specifically derivatives
  • Familiarity with the point-slope formula for linear equations
  • Knowledge of polynomial functions, particularly cubic functions
  • Ability to evaluate functions and their derivatives at specific points
NEXT STEPS
  • Study the application of derivatives in finding tangent lines for various polynomial functions
  • Explore the concept of higher-order derivatives and their geometric interpretations
  • Learn about the relationship between derivatives and curve sketching techniques
  • Investigate real-world applications of derivatives in physics and engineering
USEFUL FOR

Students learning calculus, mathematics educators, and anyone interested in the geometric interpretation of derivatives and their applications in real-world scenarios.

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!
 

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