Negative powers differentiation | Derivative rules | AP Calculus AB | Khan Academy

TL;DR

This video demonstrates how to find the derivative of a function using the power rule and evaluate it at a specific value.

Transcript

• [Voiceover] So we have the function g of x, which is equal to 2/x to the third minus 1/x squared. And what I wanna do in this video, is I wanna find what g prime of x is and then I also wanna evaluate that at x equal two, so I wanna figure that out. And I also wanna figure out what does that evaluate to when x is equal to two? So what is the slop... Read More

Key Insights

• ✊ The power rule is a fundamental tool for finding derivatives by multiplying the exponent by the coefficient and then subtracting one from the exponent.
• 😑 Rewriting expressions using basic exponent properties can simplify the process of finding derivatives.
• 🫡 The derivative of a function represents the rate of change of the function with respect to the independent variable.

Questions & Answers

Q: How do you rewrite 2/x^3 and 1/x^2 to apply the power rule?

2/x^3 can be rewritten as 2x^(-3), and 1/x^2 can be rewritten as x^(-2).

Q: What is the derivative of 2x^(-3)?

Applying the power rule, the derivative of 2x^(-3) is -6x^(-4).

Q: What is the derivative of x^(-2)?

Using the power rule, the derivative of x^(-2) is -2x^(-3).

Q: How do you evaluate g'(x) at x=2?

Plug x=2 into g'(x) = -6x^(-4) + 2x^(-3), resulting in g'(2) = -1/8.

Summary & Key Takeaways

• The video teaches how to find the derivative of the function g(x) = 2/x^3 - 1/x^2 using the power rule and basic exponent properties.

• The derivative is found to be g'(x) = -6x^(-4) + 2x^(-3).

• The video also shows how to evaluate g'(x) at x=2, resulting in the slope of the tangent line to the graph of g at x=2 being -1/8.