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## AP®︎ Calculus AB (2017 edition)

### Course: AP®︎ Calculus AB (2017 edition)>Unit 3

Lesson 3: Polynomial functions differentiation

# Differentiating polynomials review

Review your polynomial differentiation skills and use them to solve problems.

## How do I differentiate polynomials?

To differentiate polynomials, all you need are the Power rule and the basic differentiation rules. Let's differentiate 3, x, squared, minus, 2, x, plus, 1 for example.
\begin{aligned} &\phantom{=}\dfrac{d}{dx}(3x^2-2x+1) \\\\ &=3\dfrac{d}{dx}(x^2)-2\dfrac{d}{dx}(x)+\dfrac{d}{dx}(1) \\\\ &=3(2x)-2(1)+(0) \\\\ &=6x-2 \end{aligned}
Want a deeper explanation of differentiating polynomials? Check out this video.

## Practice set 1: Differentiate polynomials

Problem 1.1
• Current
f, left parenthesis, x, right parenthesis, equals, x, start superscript, 5, end superscript, plus, 2, x, cubed, minus, x, squared
f, prime, left parenthesis, x, right parenthesis, equals

Want to try more problems like this? Check out this exercise.

## Practice set 2: Evaluate polynomial derivatives

Problem 2.1
• Current
f, left parenthesis, x, right parenthesis, equals, x, start superscript, 4, end superscript, plus, 3, x, cubed, minus, x, squared
f, prime, left parenthesis, 2, right parenthesis, equals, question mark

Want to try more problems like this? Check out this exercise.

## Practice set 3: Find tangent equations

Problem 3.1
• Current
Find an equation of the line tangent to the graph of f, left parenthesis, x, right parenthesis, equals, x, start superscript, 4, end superscript, plus, 2, x, squared at the point where x, equals, 1.

Want to try more problems like this? Check out this exercise.

## Practice set 4: Find normal line equations

Problem 4.1
• Current
Find an equation of the line normal to the graph of g, left parenthesis, x, right parenthesis, equals, x, cubed, minus, 5, x, squared, plus, 10 at the point where x, equals, 2.

## Want to join the conversation?

• I think finding normals were not covered by material up to now
• In algebra and/ or geometry, we learn that a line is perpendicular (normal means perpendicular) to another if it has a slope which is opposite and reciprocal. So, just find the equation of the tangent in point slope form, and replace the tangent slope with the opposite reciprocal. I hope I could help!
• What would any applications of finding normal line equations be?
• A particle moving on a circle with constant velocity will have it's velocity expressed as the derivative of space with respect to time. It's acceleration vector will always point towards the center and will be normal to the velocity vector.
• how to find normal line equations? ........ i didn't find the content which covered this question
• u might have studied in ur class 11 that the product of the slopes of two perpendicular lines is -1 i.e. m1 x m2 = -1
(1 vote)
• Is y=mx+b a differential equation?
(1 vote)
• No. It doesn't contain any derivatives. A differential equation relates the derivatives of a function. The differential equation y'=y is satisfied by y=e^x.
• Notice that the normal line is perpendicular to the tangent line.
• i am lost from part
"We need an equation of the line that passes through (2,-2)(2,−2)left parenthesis, 2, "
​​
• In the comments above, we found out that the "normal" line is the line perpendicular to the tangent line we found. We know the formula for the tangent line. And we know the points on the tangent line. And we know that the "normal" line will have the inverse of the slope of the tangent line. So, all we have to do is re-write the tangent line formula from y = -8x+14 to y = 1/8 + m. We know the slope will get an inverse of the slope including the multiplier (-8 ) and the sign (-) that gives us a positive 1/8 slope.

If you can't see it in your mind sit down and graph it out.

What we don't have is the intercept, or m. To get it we just plug the point in that we know is on the line. This is where the tangent, and the "normal" intersect and it is the point we started working with x= 2 y = -2. We just plug them into our new "normal" line formula. -2 = 1/8(2) +m and viola! We get -2 = 1/4 + m. we solve for m by subtracting 1/4 from both sides and we get -9/4 or -2 1/4. We put that in place of m and we got it.