Understanding Derivatives: Solving with Power Rule vs. Definition

[swears]

I'm having a really hard time grabbing a hold of this definition of a derivative concept.

I know that the derivative of $$X-X^2 = 1-2X.$$, when solving w/ the power rule.
But, I get really lost when I need to solve it using the definition of the derivitave. Can someone please explain to me how I get from the 1st step to the 2nd.

$$F(X) = X - X^2$$

1. $$F'(X) = \frac{F(X+H) - F(X)}{H}$$

2. $$= \frac{(X+H)^2 - X^2}{H}$$The way I tried it, I just input $$X - X^2$$ for X and I got $$(X - X^2 + H) - (X-X^2)$$

 

[vsage]

I'm having a really hard time grabbing a hold of this definition of a derivative concept.

I know that the derivative of $$X-X^2 = 1-X.$$, when solving w/ the power rule.

Nope... try to do that derivative again

The way I tried it, I just input $$X - X^2$$ for X and I got $$(X - X^2 + H) - (X-X^2)$$

If f(x) is x-x^2, what is f(x+h) it's not x-x^2 + h, I will tell you that much, but it sounds like you have some issues with algebra as well. You applied the definition wrong, so give it another go.

 

[swears]

I fixed that first part sorry. lol

 

[swears]

If f(x) is x-x^2, what is f(x+h) it's not x-x^2 + h,

Yeah, that's my problem, I don't know how to do that.

 

[Jameson]

Split up your function to make things much simpler. Let's say you have $$f(x)=x$$ and $$g(x)=-x^2$$

$$f'(x)=\lim_{h \to 0}\frac{(x+h)-(x)}{h}$$

$$g'(x)=\lim_{h \to 0}\frac{-(x+h)^2-(-x^2)}{h}$$

Work from there.

 

[Jameson]

Reading your first post, I'll try to clarify what I think is your problem. If you have $$f(x)=x^2$$, then something like f(2) is easy, right? You just plug in 2 for x and get 4. But what is f(a)? Do the same thing. f(a)=a^2. What is f(x+h) then? Plug in (x+h) for x. f(x+h)=(x+h)^2

 

[swears]

$$f'(x)=\lim_{h \to 0}\frac{(x+h)-(x)}{h}$$

$$g'(x)=\lim_{h \to 0}\frac{-(x+h)^2-(-x^2)}{h}$$

Why does the H get squared and negated?

 

[vsage]

I fixed that first part sorry. lol

It's still not fixed on my screen. the derivative of x-x^2 should be 1-2x (I see it in the TeX code but for some reason it isn't showing up on your post).

Let me elaborate a little. If f(x) = x-x^2, then f(x+h) = (x+h) - (x+h)^2, so expand this out. What is then f(x+h) - f(x)?

 

[swears]

It's still not fixed on my screen. the derivative of x-x^2 should be 1-2x (I see it in the TeX code but for some reason it isn't showing up on your post)

try refresh F5

 

[swears]

Plug in (x+h) for x.

So I'm doing it backwards here?

 

[Jameson]

I don't understand what you mean by backwards. Look at my original post to see the correct method of using the limit definition of a derivative. I'm sorry I can't be more helpful right now.

 

[swears]

Plug in (x+h) for x.

Ok, if i try it this way, I get: $$(x+h) + h - (x+h)^2$$

 

[vsage]

You want to solve the limit of $$\frac { f(x+h) - f(x)} {h}$$ as h approaches 0, right? If f(x) = x - x^2, then f(x+h) - f(x) = (x+h) - (x+h)^2 - (x - x^2)). I'm a little concerned though as these problems you are having are algebraic and not calculus-based, as your thread suggests. Applying basic rules of algebra (up to factoring in the denominator to the expression) should fetch you the right answer.

 

[swears]

You wouldn't happen to have a link for these algebra rules, would you?

I understand how you plugged in $$x-x^2$$ for f(x).
but,
I don' see how you got $$(x+h) - (x+h)^2$$ for $$f(x+h)$$ though.

 

[FrogPad]

I'm going to repeat what was said here. But maybe seeing it worked out will help you a little bit more.

You know the definition of the derivatie is:
$$f'(x)=\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}$$

Now you seem to be caught up with the algebra of functions. Let me go through a few examples with you:

Lets say you have a function:
$$f(x)=x^2$$

Now let's plug in a "cat" :)

$$f(cat)=f(x=cat)=(cat)^2$$

Now let's plug in $a+b$

$$f(a+b)=f(x=a+b)=(a+b)^2$$


Now plugging in a cat is really not any harder then plugging what you need into the limit definition I gave above. But let's try an example:

Lets create a function, say:
$$f(x)=x^{1000}$$

You seem to be able to use the "non-limit rules" to find the derivative, so you know that:
$$f'(x)=1000x^{999}$$

Right? That's pretty easy right.

Now let's use the limit definition.

$$f'(x)=\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h} = \frac{(x+h)^{1000}-x^{1000}}{h}$$

Notice what we did.

We just plugged in our values.

Do you see that:
$$f(x+h)=(x+h)^{1000}$$

You can think of it like this if you want.
let [itex]x+h = cat [/tex]<br /> $$f(x)=x^{1000}$$<br /> $$f(cat)=(cat)^{1000}$$<br /> <br /> And we know that $cat = x+h$ so...<br /> $$f(cat=x+h)=(cat=x+h)^{1000}$$<br /> or in a more popular way to write it:<br /> $$f(x+h)=(x+h)^{1000}$$<br /> <br /> So really, you are just plugging in values.[/itex]

 

[swears]

Thanks for your help.

So I did: $$f(x) = x - x^2$$

$$f'(x) = \frac {x + h - (x + h)^2 - x - x^2}{h}$$

Now I tried to simplify but I get stuck or have done it wrong.

$$= \frac {x + h - x^2 + 2xh + h^2 - x - x^2}{h}$$

$$= \frac {h - 2x^2 + 2xh + h^2}{h}$$ // I canceled out the x and combined the x squared

Is this right?

 

[TD]

Thanks for your help.

So I did: $$f(x) = x - x^2$$

$$f'(x) = \frac {x + h - (x + h)^2 - x - x^2}{h}$$

Watch out, at the end of the numerator you should have "-f(x)". With f(x) = x-x², this becomes -(x-x²) = -x+x².

 

[swears]

So, your saying I need to distribute the minus sign to f(x)?

 

[TD]

Of course, when it says -f(x), f(x) is considered as a whole and should therefore be placed within parathesis. This gives -(x-x²) which simplifies to -x+x² and not -x-x².

 

[swears]

Of course, when it says -f(x), f(x) is considered as a whole and should therefore be placed within parathesis. This gives -(x-x²) which simplifies to -x+x² and not -x-x².

Ah ok, thanks for explaining that.

 

[swears]

Tried again,

$$f(x) = x - x^2$$

$$f'(x) = \frac {x+h - (x + h)^2 - (x - x^2)}{h}$$

$$= \frac {x + h - x^2 + h^2 + 2xh - x + x^2}{h}$$

$$= \frac {h + h^2 + 2xh}{h}$$

$$= \frac {h [2x + h]}{h}$$

$$= 2x + h$$ // Now I'm lost

 

[TD]

You really have to watch our signs, there's a minus before (x+h)² !

 

[FrogPad]

Try not to skip steps when you are getting used to this stuff.

So if you have:

$$f(x) = x-x^2$$
$$f'(x)=\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}$$

Use paranthesis! Even if it is really obvious. At least until you are comfortable with order of operations.

so...

$$f'(x)=\lim_{h\rightarrow 0} \frac{((x+h)-(x+h)^2)-(x-x^2)}{h}$$

Now here's an example for you to remember. Recall that a number multiplied by 1 is the same number.
$$(1)(2) = 2$$

Now what is this?
$$-(2) = 2$$?

What about this?
$$-(x^2-2+3) =?$$

You can think about the negative sign as $-1$. So in the first example we can rewrite it as:
$$(-1)(2) = -2$$

and the second example:
$$(-1)(x^2-2+3)$$

Do what's in the parenthesis first:
$$(-1)(x^2-2+3)=(-1)(x^2+1)$$
$$(-1)(x^2+1) = (-1)(x^2) + (-1)(1) = -x^2 - 1$$


So back to this example:
$$f'(x)=\lim_{h\rightarrow 0} \frac{((x+h)-(x+h)^2)-(x-x^2)}{h}$$

You can also think of this like:
$$f'(x)=\lim_{h\rightarrow 0} \frac{((x+h)+(-1)(x+h)^2)+(-1)(x-x^2)}{h}$$

Also, if it gets to confusing using just parenthesis. Remember that you can use other symbols too.

For example:
$$f'(x)=\lim_{h\rightarrow 0} \frac{[(x+h)+(-1)(x+h)^2]+[(-1)(x-x^2)]}{h}$$

Notice how the terms in between [...] are used to represent the function.

For example:
$$f(x) = x-x^2$$
$$g(x) = f(a+b)^2 + f(\lambda)$$

Now if you use the brackets, it is easy to see:
$$g(x) = [(a+b)-(a+b)^2]^2 + [\lambda -\lambda^2]$$

 

[swears]

Ok, I fixed my mistake and I got :

$$\frac {h(-2x-h)}{h}$$

So, now do I need to set -2x-h = 0?

Because when I do that I get -2x = h . But I know the derivative is 1-2x. How do I get the 1?

 

[TD]

How did you get that? Let's see:

$$\mathop {\lim }\limits_{h \to 0} \frac{{\left( {\left( {x + h} \right) - \left( {x + h} \right)^2 } \right) - \left( {x - x^2 } \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{x + h - x^2 - 2xh - h^2 - x + x^2 }}{h}$$

Now simplying and factoring h gives:

$$\mathop {\lim }\limits_{h \to 0} \frac{{h - 2xh - h^2 }}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{h\left( {1 - 2x - h} \right)}}{h}$$

Which is a bit different from your answer. Check it again.
Then, cancel out the h and let h go to 0.

 

[swears]

Oh yeah, I forgot the first h.

$$1-2x-h = 0$$

$$1-2x = h$$

So I'm guessing the h stands for the derivative. I'm going to try a few more of these on my own.

I'd like to thank everyone for helping me, I couldn't have done it w/o you. Looks like I won't be dropping calculus yet.

 

[TD]

No, it's not an equation (where did the = 0 come from?) It was a limit of that expression, so in "1-2x-h", let h go to 0. What do you get?

 

[swears]

oh, so I plug in 0 for h.

which gets 1-2x-0 or 1-2x

 

[TD]

Indeed, and is that compatible with the answer you'd find by using the normal derivative rules?

 

[swears]

Yes, with the power rule.