Product rule(help)

[stopandthink]

I'm reading the intro to calculus on the forum and i'm confused...

Basically it states [math]f(x) = (x - 2)(x + 4)[/math]

 

and to find it's derivative you basically have to add the product rule formula, but

i got completely lost at the last step,which is is where i need help at, and by the way would it not have been easier just multiplying the function

at the very start... I mean i see that in the last step all that was done was just flipping it and then multiplying..

Edited October 13, 2012 by stopandthink

[Cap'n Refsmmat]

Could you be more specific? What happens in the last step that confuses you?

 

It would indeed be easier at the start to multiply things out, but often times you can't do that -- what if [math]f(x) = x\sin x[/math] or [math]f(x) = x e^x[/math]?

[stopandthink]

Could you be more specific? What happens in the last step that confuses you?

Of course.

I got everything up to this point

[math]1 \cdot (x + 4) + 1 \cdot (x - 2) = (x + 4) + (x - 2) = 2x + 2[/math]

1*(x+4)+1*(x-2)

Where in the formula does it say to multiply by a one.

It would indeed be easier at the start to multiply things out, but often times you can't do that -- what if [math]f(x) = x\sin x[/math] or [math]f(x) = x e^x[/math]?

I'm not that far into calculus to understand these type of functions yet, but i will.

 

-------

 

it basically left me hanging after

"And since we can find the derivative of things like (x - 2"

after that im lost

Edited October 13, 2012 by stopandthink

[Cap'n Refsmmat]

The formula for the derivative of [math]uv[/math] says to find [math]u'v + v'u[/math]. So the 1s come from taking the derivative of each part -- the derivatives of [math](x+4)[/math] and [math](x-2)[/math] are both 1, yes?

[stopandthink]

I didn't know you could apply [math]\frac{d}{dx} ax^n = anx^{n - 1}[/math]

To [math](x-2)[/math] or [math](x+4)[/math]

 

______

Like this

(x-2)^1 = 1*(x-2)

is that what you did?

Edited October 13, 2012 by stopandthink

[Cap'n Refsmmat]

What is the derivative of [math](x-2)[/math]?

 

Do it in parts. The derivative of x is 1, the derivative of -2 is 0. So the derivative is 1 + 0 = 1.

[stopandthink]

What is the derivative of [math](x-2)[/math]?

 

Do it in parts. The derivative of x is 1, the derivative of -2 is 0. So the derivative is 1 + 0 = 1.

 

Great that cleared up what i was having difficulty with. Thank you!

[ACUV]

I didn't know you could apply [math]\frac{d}{dx} ax^n = anx^{n - 1}[/math]

To [math](x-2)[/math] or [math](x+4)[/math]

 

______

Like this

(x-2)^1 = 1*(x-2)

is that what you did?

 

 

I think you could apply this rule, you are applying it wrong though here. Here is how I would apply this rule to the situation.

 

the derivative of ( x - 2 ) ^ 1 = ( 1 ) ( ( x - 2 ) ^ 0 ) ( d/dx ( x - 2 ) ) = ( 1 ) ( 1 ) ( 1 ) = 1

 

It makes the process longer in this situation but it is correct.