stopandthink

I see now. thaks

How do i find the square root of a negative number....?

Lets say [math] \sqrt -4 [/math]

My calculator gives me the answer 0

is it because [math] -a(-a)=+b [/math] where a equals -2(i assume -2 should be its squared root but its not)

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!

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?

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.

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it basically left me hanging after

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

after that im lost

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..

Lol! i see what i did wrong,

Ignore this post! Thanks.

Look at how you are expanding your brackets of (1+h)^2.

It is the same as saying (1+h)(1+h)... From here you should get (1+ 2h + h^2)

 

I dont know where you are making your mistake so if you could show what you did then that could be helpful. I am pretty certain though that you are not expanding your brackets and multiplying out correctly. Remember from (1 + h)(1 + h) you must multiply every term in the first bracket by every term in the second bracket: i.e. (1*1 + 1*h + h*1 + h*h). I think this was called FOIL.. First outter inner last or something like that. If you are still struggling look up bracket expansion and how to multiply out of brackets.

 

(x+1)^3 = (x+1)(x+1)(x+1) = (x+1)(x^2+2x+1)= ((x^3)+(2x^2)+(x) + (x^2)+)(2x)+(1))

 

try expand this: (x+5)^2

Thank you! i was stuck trying to work it out some other way... Lesson learned.

I can't figure out how (1+h)^2 -1 = 1 + 2h+ h^2-1

my answer is 2+ 2h^2-1

Of course, my answer is incorrect but i cant see where it is i'm making a mistake.

[math]\frac{(1+h)^2 - 1}{h} = \frac{1 + 2h + h^2 - 1}{h} = \frac{2h + h^2}{h} = \frac{h(2 + h)}{h} = 2 + h[/math]

I'm currently using one of the forums tutorial on calculus... And i'm stuck with a problem.

The problem is i know it works out because iv'e substituted "h" with an actual integer, and my dilemma is i can't figure out how it went from, [math]\frac{(1+h)^2 - 1}{h} = \frac{1 + 2h + h^2 - 1}{h}[/math]

Ignore the /font font=monospace thing, can't figure out how to remove it

Because one sees the other guy's face dirty and supposes his is also, but the other guy sees the clean face and assumes his face is also clean... i haven't looked at the comments so im sorry if this has already been answered.