**Product Rule**for derivatives and as usual we had white board questions!

**1. f(x) = x**

^{3}+ 2x^{2}- 8x + 1Find the equation of the tangent line at

**x=2**

f(2) = 1

f'(x) = 3x

^{2}+ 4x - 8

f'(2) = 3(2)

^{2}+ 4(2) - 8

= 3(4) + 8 - 8

= 12 + 8 - 8

= 12

**equation: y - 1 = 12(x-2)**

**2. g(x) = x**

^{3/2}- x^{1/2}Find the equation of the tangent line

**x=4**

g(4) = 6

g'(x) = (3/2)x - (1/2)x

g'(4) = (3/2)x

^{1/2}- (1/2)

^{-1\2}

= (3/2)(4)

^{1/2}- (1/2)

^{-1\2}

= 11/4

**equation: y - 6 = 11/4(x-4)**

**3. Multiply**

(x + 3)(2x +5)

(x + 3)(2x +5)

I know that everyone can easily do this problem but this is going to help us derive the product derivative rule. I wasn't taught how to multiply by using "the area model of multiplication" and it probably wouldn't have changed how i multiplied but it will serve its purpose in the next question. Mr.K showed us an example with

**(x + 3)(2x +5)**

**= 2x**

^{2}+ 11x + 15With this example, it helped me understand how to find the derivative rule of

**F(x)=f(x) g(x).**

**Area I: [g(x+h) - g(x)]f(x)**

**Area II: [f(x+h)-f(x)]g(x)**

**Area III: [f(x+h)-f(x)][g(x+h)-g(x)]**

With these areas, we were able to derive the product rule of the derivative.

**The Product Rule of the Derivative is f(x)g'(x) + g(x)f'(x).**

No Homework today. My work here is done. Jann, I choose you!

## 2 comments:

I loved this post. I teach my students the area model in Grade 8. I now have another reason for them to learn this model. Thanks for doing such a great job. YOur use of colour and images is beyond excellence. I nominate you for the Hall of Fame. I hope your classmates read this.

Mr. Harbeck

Sargent Park School

Hi Lindsay! This scribe is a work of art. Not only that it's really good, its very helpful and I think you covered everything we did that day. I nominate you for the Hall of Fame! XD

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