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## AP®︎/College Calculus BC

### Course: AP®︎/College Calculus BC>Unit 2

Lesson 7: Derivative rules: constant, sum, difference, and constant multiple: connecting with the power rule

# Differentiating polynomials

AP.CALC:
FUN‑3 (EU)
,
FUN‑3.A (LO)
,
FUN‑3.A.3 (EK)
Sal differentiates f(x)=x⁵+2x³-x², and evaluates the derivative at x=2.

## Want to join the conversation?

• What do you do when you have a function like f(x)=4x^2+2x? I know that for the first part (4x^2) you can use the power rule, but what about the second part (2x)?
• Basically - x^1 becomes x^0, which becomes 1. Any constant times 1 becomes that constant. Therefore f'(x) of 2x = 2 * x^(1-1) = 2*1 = 2. I just like to think that x^1 "disappears" when derived, leaving just the constant in front of it.
• In this video and in others, Sal takes the derivative of both sides of the equation. Is that always “legal”, mathematically speaking?
• Yes, if two functions are equal, then their derivatives are equal.
• How do you take derivatives with respect to y?
• When Sal writes f(x), he just uses x as a variable, and he means f(a variable). Therefore, you could substitute f(x) for f(y), and get the same graph. As to actually taking the variable, you would do d*f(y)/dy, or take the limit as h approaches 0 of [f(y+h) - f(y)]/h
• How can we find minimum/maximum values of polynomials by differentiation?
• Given that with the Derivative we are able to get the Slope of tangent lines to our function at any x values, if we set our Derivative expression equal to 0 we are going to find at what x values we have the Slope of our tangent line equaling 0, which would be just a horizontal line.

The only time that happens is at min/max values. We can only get horizontal tangent lines at those values, if you think about it. With the rest of our function our tangent lines have Slopes that are either decreasing or increasing along the function.

We call these x values Critical numbers. Where we have horizontal tangent lines.

To determine whether we have a min/max, we would test other x values around the Critical points we already found into our Derivative function (evaluating it), to see if our Slope is positive (that would imply that the function is increasing on that interval), or if it is negative (our function would be decreasing).

If there's a change from having a positive Slope (our function increasing) to have a negative Slope (our function decreasing), then that is a Relative/Local maximum. If there's a change from decreasing to increasing, then that is a Relative/Local minimum.

Finally, we would evaluate our Critical numbers into our original function to actually figure out the points of our mins/maxes.

The process of doing this, finding Critical numbers by setting our Derivative equal to 0, and testing values around them to determine whether we have a min/max, is known as the First Derivative Test.
• What would be the answer for
2x^2 + 30x = 0
Value for x
We can calculate it as 15 by using simple equation formula. But what about derivative method
• Not totally sure about what you want to convey...The solutions to that parabola are 0 and -15 and its derivative is 4x +30 by using the power rule.....the slope of this linear function is 4.....
• that's some funky notation d/dx. you'd think the d's would cancel out.
• 'dx' is a single symbol, and refers to the differential of x. Think of it as an infinitesimally small amount of x. It is not the product of d and x.
• I learned this but I cant differentiate 2^x. How can I do it ?
• 2^x is exponential, so you can't differentiate it like a polynomial. d/dx[2^x] = ln(2) * 2^x.
• What do you do when you have a function like f(x)=4x^2+2x? I know that for the first part (4x^2) you can use the power rule, but what about the second part (2x)?
• You can still use the power rule on the 2x (x is still raised to the 1 power so the derivative of 2x is 2). Using the power rule you multiply the power of x which in the case of 2x is 1 and 2*1 yields 2 then subtract 1 from the power of x you get x^0 which is equal to 1. You end up with 2*1 which is 2. The derivative of 2x is 2.
• @ Why is the derivative of polynomial is the sum of the difference of derivative of each item? It is "given" in this video, but I want to understand "why".
• Let f(x) and g(x) be two functions. From the definition of a derivative:

lim(h-->0)f(x) = f(x+h)-f(x)/h

lim(h-->0)g(x) = g(x+h)-g(x)/h

Now, add the limits. You'll get lim(h-->0)(f(x) + g(x)) = (f(x+h)-f(x)+g(x+h)-g(x))/h

Now, let a function a(x) exist such that a(x) = f(x) + g(x). So, lim(h-->0)a(x) = a(x+h)-a(x)/h

As h(x) is just f(x) + g(x), we have:

lim(h-->0)(f(x)+g(x)) = (f(x+h)-f(x)+g(x+h)-g(x))/h

This clearly shows that derivative of the sum/difference = sum/difference of the derivatives.