How do I find the gradient of the tangent at a given point on a quadratic curve?

[adjacent]

Homework Statement

Find the gradient of the tangent at x on the following curve
##y=3x^2##

Homework Equations

$$\lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$$

The Attempt at a Solution

I know that it's ##6x##.
$$\frac{3(x+\Delta x)^2-3x^2}{\Delta x}$$
$$=\frac{3x^2+6x\Delta x +3{\Delta x}^2-3x^2}{\Delta x}$$
$$=\frac{3\Delta x(2x + \Delta x)}{\Delta x}=2x+3\Delta x$$
$$\lim_{\Delta x\rightarrow 0}2x+3\Delta x =2x$$


Where did I go wrong?

 

[Curious3141]

$$=\frac{3\Delta x(2x + \Delta x)}{\Delta x}=2x+3\Delta x$$

In this step, have you distributed the 3 correctly on multiplying?

 

[adjacent]

In this step, have you distributed the 3 correctly on multiplying?

Ohhhh

$$=\frac{3\Delta x(2x + \Delta x)}{\Delta x}$$
$$=2\Delta x(2x+\Delta x)$$
$$\lim_{\Delta x \to 0}2\Delta x(2x+\Delta x)=0$$

Whaat?

 

[adjacent]

I just realized that I made an algebra mistake.
$$\frac{3\Delta x(2x + \Delta x)}{\Delta x}=3(2x+\Delta x)$$
$$\lim_{\Delta x \to 0}6x+ \Delta x=6x$$


How foolish I am!

 

[Curious3141]

I just realized that I made an algebra mistake.
$$\frac{3\Delta x(2x + \Delta x)}{\Delta x}=3(2x+\Delta x)$$
$$\lim_{\Delta x \to 0}6x+ \Delta x=6x$$


How foolish I am!

Believe me, I've been far more foolish on many occasions! Just a simple error, nothing to beat yourself up about.

 

[adjacent]

Thanks.
Another question came to mind,
Differentiate ##f(x)=x^2+3x+1##
I think the correct answer would be ##2x+3##

But I have a doubt.
I used the addition rule there , ##(v+u)'=v'+u'##
So what do I do about the derivative of 1? wouldn't it be ∞?

 

[Curious3141]

Thanks.
Another question came to mind,
Differentiate ##f(x)=x^2+3x+1##
I think the correct answer would be ##2x+3##

But I have a doubt.
I used the addition rule there , ##(v+u)'=v'+u'##
So what do I do about the derivative of 1? wouldn't it be ∞?

Why would it be infinite? The derivative is basically a rate of change. What is the rate of change of a constant function?

 

[adjacent]

Why would it be infinite? The derivative is basically a rate of change. What is the rate of change of a constant function?

Oh. I thought it was x=1.

Actually I am self-learning calculus.That's why I make these simple mistakes.

Another question.

##(u+v)'=u'+v'##.
So what are u and v here? Are they functions?
In the function ##f(x)=x^2+3x+1##, is ##x^2##,##3x## and ##1## considered functions?Aren't they a part of the function ##f##?

 

[SteamKing]

Oh. I thought it was x=1.

Actually I am self-learning calculus.That's why I make these simple mistakes.

Another question.

##(u+v)'=u'+v'##.
So what are u and v here? Are they functions?
In the function ##f(x)=x^2+3x+1##, is ##x^2##,##3x## and ##1## considered functions?Aren't they a part of the function ##f##?

There's nothing that says f can't be the sum or product of other functions.

 

[adjacent]

So ##x^2+3x+1## is treated like sum of functions?

Q.
Why is ##(\frac{1}{v})'=\frac{v'}{v^2}##? I saw it from here

If I apply the power rule, I get
##v^{-1}=-v^{-1-1}=-v^{-2}=-\frac{1}{v^2}##

Does that mean that rule is wrong?

 

[Panphobia]

it said
-f'/f^2
so it is right, v' is just the derivative of the denominator and not the reciprocal, in your example, let's use that.
v' = 1
so (1/v)' = -1/v^2

 

[adjacent]

it said
-f'/f^2
so it is right, v' is just the derivative of the denominator and not the reciprocal, in your example, let's use that.
v' = 1
so (1/v)' = -1/v^2

ohh. So it's v itself!
So what's the use of that rule? It can be easily derived from the power rule,right?

EDIT:Using the quotient rule is easier.
I just tried using the power rule for
##\frac{1}{2x^2+3x}## and it gave me ##-\frac{1}{(2x^2+3x)^2}## instead of ##-\frac{4x+3}{(2x^2+3x)^2}## which I got by using both the reciprocal and the quotient rule.

 

[Panphobia]

Yea, I wouldn't suggest memorizing anything more than the chain rule, power rule, and product rule. That is all you need for differentiating, of course there is the quotient rule, but that can be derived with the product rule + chain rule + power rule. The sum and difference rules I never learned explicitly, they were just implied to be true.

 

[SteamKing]

So ##x^2+3x+1## is treated like sum of functions?

Yes. The derivative of the sum of several different functions is equal to the sum of the derivatives of each function. For example:

g(x) = f1(x) + f2(x) + f3(x), where f1, f2, and f3 are all continuous functions of x.

g'(x) = f1'(x) + f2'(x) + f3'(x), where the ' indicates the derivative w.r.t. x.

This is also analogous to the limit of a sum being equal to the sum of the limits, so long as each limit exists.

 

[adjacent]

Yea, I wouldn't suggest memorizing anything more than the chain rule, power rule, and product rule. That is all you need for differentiating, of course there is the quotient rule, but that can be derived with the product rule + chain rule + power rule. The sum and difference rules I never learned explicitly, they were just implied to be true.

What about those trigonometric things and exponents etc?

Note: I have edited my last post there,please see it.

 

[Panphobia]

Oh I meant rules, but yea the actual derivatives you will need to memorize, like (sin(x))' = cos(x) and (ln(x))' = 1/x, and for your edited post, so imagine
v = (2x^2 + 3x)^(-1)
then
v' = -4x -3/(2x^2+3x)^(2)
You get it now?
Have you learned the chain rule yet? Usually I find that it is what most people have trouble with when learning differential calculus.

 

[adjacent]

v = (2x^2 + 3x)^(-1)
then
v' = -4x -3/(2x^2+3x)^(2)
You get it now?

No.
See what I did:
$$\frac{d}{dx}(2x^2+3x)^{-1}=-1(2x^2+3x)^{-1-1}=-1(2x^2+3x)^{-2}$$
$$=-\frac{1}{(2x^2+3x)^2}$$

What is the wrong step here?

Have you learned the chain rule yet? Usually I find that it is what most people have trouble with when learning differential calculus.

I will learn it today.

 

[HallsofIvy]

That is a difficulty with using the "Newton notation" for the derivative (primes) rather than the Leibniz notation- you don't make the variable explicit. It is true that [itex]dx^{-1}/dx=-x^{-2}[/itex]. And [itex]dy^{-1}/dy= -y^{-2}[/itex] and [itex]du^{-1}/du= -u^{-2}[/itex] and [itex]dv^{-1}/dv= -v^{-2}[/itex]. It doesn't matter what you call the variable!

So if [itex]v= 2x^2+ 3x[/itex] then [itex](2x^2+ 3x)^{-1}= v^{-1}[/itex] so that
[tex]\frac{d(2x^2+ 3x)^{-1}}{dv}= \frac{dv^{-1}}{dv}= -v^{-2}= -(2x^2+ 3x)^{-2}[/tex]

But that is the derivative with respect to v (or with respect to [itex]2x^2+ 3x[/itex]), not with respect to x!
To do that you can use the chain rule:
[tex]\frac{d f(v)}{dx}= \left(\frac{df(v)}{dv}\right)\left(\frac{dv}{dx}\right)[/tex]

 

[Curious3141]

No.
See what I did:
$$\frac{d}{dx}(2x^2+3x)^{-1}=-1(2x^2+3x)^{-1-1}=-1(2x^2+3x)^{-2}$$
$$=-\frac{1}{(2x^2+3x)^2}$$

What is the wrong step here?


I will learn it today.

You didn't apply chain rule here. You just differentiated treating ##2x^2+3x## as if it was x.

Chain rule allows differentiation of composite functions. So if you have a composite function like ##h(x) = f(g(x))## and you want to find the derivative, it's given by ##h'(x) = f'(g(x)).g'(x)##. Note that the dot is often used to repent multiplication in unwieldy algebraic expressions.

In your case ##f(x) = x^{-1}## while ##g(x) = 2x^2 + 3x##. Think about why this is so. You need to have a great grasp of functional theory before going on to calculus.

The easiest albeit informal way I can describe the intuitive application of chain rule is as "treat function as a function of another part, then differentiate the function as if it were just a function of x BUT subbing the part in place of x and THEN multiply by the derivative of the part." This is how I think about Chain Rule when I actually apply it. You can iterate the rule as many time as you like and you'll see this when you read more and have to deal with nested functions like ##\sin^3(\cos(x))##.

 

[adjacent]

But that is the derivative with respect to v (or with respect to [itex]2x^2+ 3x[/itex]), not with respect to x!

You didn't apply chain rule here. You just differentiated treating ##2x^2+3x## as if it was x.

So If I use the quotient formula (##(\frac{u}{v})'=\frac{u'v-uv'}{v^2}##), am I differentiation with respect to x?
I don't think so.Why are we not applying chain rule in this case?
Aren't we considering v as the denominator(##2x^2+ 3x##)?

##\sin^3(\cos(x))##.

I have learned chain rule today.
The chain rule is ##(fg(x))'=f'g(x) . g'(x)##
How do I extend this to more than two functions?
The lecture videos I watched did not explain chain rules well! But he said that if you know chain rule,you can conquer the world!(lol, I don't know why.Is this so useful?)

 

[adjacent]

The chain rule is ##(fg(x))'=f'g(x) . g'(x)##
How do I extend this to more than two functions?

I think I understand it now.

What we do is first take the derivative of the outermost function and plugin in all the inner functions in it then multiply the derivative of that inner function's outer function.. Then repeat the process.Is that all?
I see now why its called "chain rule".I think I will need to draw a chain to remember this

 

[Ray Vickson]

So If I use the quotient formula (##(\frac{u}{v})'=\frac{u'v-uv'}{v^2}##), am I differentiation with respect to x?
I don't think so.Why are we not applying chain rule in this case?
Aren't we considering v as the denominator(##2x^2+ 3x##)?

I have learned chain rule today.
The chain rule is ##(fg(x))'=f'g(x) . g'(x)##
How do I extend this to more than two functions?
The lecture videos I watched did not explain chain rules well! But he said that if you know chain rule,you can conquer the world!(lol, I don't know why.Is this so useful?)

Why do you say "I don't think so"? The quotient rule--written out in full--says that
[tex] \frac{d}{dx}\left( \frac{u(x)}{v(x)} \right)
= \frac{u'(x) v(x) - u(x)v'(x)}{v(x)^2}[/tex]
What is there about this that you think is wrong, or did I misunderstand what you were trying to say?

 

[adjacent]

Why do you say "I don't think so"? The quotient rule--written out in full--says that
[tex] \frac{d}{dx}\left( \frac{u(x)}{v(x)} \right)
= \frac{u'(x) v(x) - u(x)v'(x)}{v(x)^2}[/tex]
What is there about this that you think is wrong, or did I misunderstand what you were trying to say?

What I mean to say is this:
the quotient rule is this:
$$\frac{d}{dx}\left( \frac{u(x)}{v(x)} \right)$$

Are not we considering u and v like they were just a single function?
v is a composite function so why shouldn't we apply chain rule here?

 

[Ray Vickson]

What I mean to say is this:
the quotient rule is this:
$$\frac{d}{dx}\left( \frac{u(x)}{v(x)} \right)$$

Are not we considering u and v like they were just a single function?
v is a composite function so why shouldn't we apply chain rule here?

Well, if ##v(x) = w[g(x)]## you need to compute ##v'(x)## using the chain rule on ##w## and ##g##. However, that just tells you what you need to do to find ##v'##; it does not in any way change the quotient rule for ##u/v##. In fact, nothing stops both ##u## and ##v## from being composite, so you would need to use the chain rule to get both ##u'## and ##v'##. However, the quotient rule would still be 100% correct exactly as written; it would just take a lot more work to get both ##u'## and ##v'##.

 

[adjacent]

Well, if ##v(x) = w[g(x)]## you need to compute ##v'(x)## using the chain rule on ##w## and ##g##. However, that just tells you what you need to do to find ##v'##; it does not in any way change the quotient rule for ##u/v##. In fact, nothing stops both ##u## and ##v## from being composite, so you would need to use the chain rule to get both ##u'## and ##v'##. However, the quotient rule would still be 100% correct exactly as written; it would just take a lot more work to get both ##u'## and ##v'##.

Really?
See my workings for ##\frac{1}{2x^+3x}##
Applying the quotient rule,I get:
##\frac{0*(2x^+3x)-1*(4x+3)}{(2x^2+3x)^2}##
$$=-\frac{4x+3}{(2x^2+3x)^2}$$

I didn't use any chain rule here.
Then why should my answer using power rule be wrong?(I think I didn't really understand what HallsofIvy and Curious cat said)
Here it is:
##\frac{1}{2x^2+3x}=(2x^2+3x)^{-1}##
$$\frac{d}{dx}(2x^2+3x)^{-1}=-1(2x^2+3x)^{-1-1}=-1(2x^2+3x)^{-2}$$
$$=-\frac{1}{(2x^2+3x)^2}$$

 

[Ray Vickson]

Really?
See my workings for ##\frac{1}{2x^+3x}##
Applying the quotient rule,I get:
##\frac{0*(2x^+3x)-1*(4x+3)}{(2x^2+3x)^2}##
$$=-\frac{4x+3}{(2x^2+3x)^2}$$

I didn't use any chain rule here.
Then why should my answer using power rule be wrong?(I think I didn't really understand what HallsofIvy and Curious cat said)
Here it is:
##\frac{1}{2x^2+3x}=(2x^2+3x)^{-1}##
$$\frac{d}{dx}(2x^2+3x)^{-1}=-1(2x^2+3x)^{-1-1}=-1(2x^2+3x)^{-2}$$
$$=-\frac{1}{(2x^2+3x)^2}$$

You need the chain rule:
[tex] \frac{d}{dx} (2x^2 + 3x)^{-1} = (-1)(2x^2 + 3x)^{-2} \cdot \frac{d}{dx} (2x^2 + 3x) = -\frac{4x + 3}{(2x^2+3x)^2}[/tex]
There are no contradictions; you are confusing yourself by making errors. You really must be careful if you want to learn calculus successfully. Carelessness is deadly.

 

[adjacent]

You need the chain rule:
[tex] \frac{d}{dx} (2x^2 + 3x)^{-1} = (-1)(2x^2 + 3x)^{-2} \cdot \frac{d}{dx} (2x^2 + 3x) = -\frac{4x + 3}{(2x^2+3x)^2}[/tex]
There are no contradictions; you are confusing yourself by making errors. You really must be careful if you want to learn calculus successfully. Carelessness is deadly.

hmm. Yeah.I got it.

All this confusions were because I thought ##\frac{1}{2x^2+3x}## and applying the quotient rule is same as ##(2x^2+3x)^{-2}## and applying the power rule.

Thanks for the help.I really appreciate it.
Thanks everyone!