Derivative of a Constant: The Quotient Rule

[adjacent]

Homework Statement

$$\frac{\text{d}}{\text{d}x}(\frac{s}{x})$$
Quotient rule : $\frac{u'v-uv'}{v^2}$
So $\frac{1x-s1}{x^2}$
$=\frac{-s}{x}$
Wolfram says its $\frac{-s}{x^2}$

 

[BOAS]

Homework Statement

$$\frac{\text{d}}{\text{d}x}(\frac{s}{x})$$
Quotient rule : $\frac{u'v-uv'}{v^2}$
So $\frac{1x-s1}{x^2}$
$=\frac{-s}{x}$
Wolfram says its $\frac{-s}{x^2}$

\frac{s}{x} = sx^{-1}

The quotient rule is not necessary because the numerator is not a function of x

 

[Fightfish]

In your application of the quotient rule, you seemed to have taken \frac{d}{dx} s = 1, which is not true.

 

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\frac{s}{x} = sx^{-1}

The quotient rule is not necessary because the numerator is not a function of x

Under which circumstances should I use the quotient rule then?

 

[Fightfish]

There is nothing special about the quotient rule; it is derived from the product rule.
\frac{d}{dx} \left(\frac{u}{v}\right) = \frac{du}{dx} \left(\frac{1}{v}\right) + u \frac{d}{dx}\left(\frac{1}{v}\right) = \frac{u'v - uv'}{v^{2}}
The product rule is more general and does not require manipulation into a fractional form to apply, so personally I rarely use the quotient rule. Of course, sometimes, the quotient rule can be slightly faster, but I stress that they are equivalent methods.

In the problem that you posed, it happens that the product rule is slightly more straightforward, but if you want to apply the quotient rule, that is fine too, and should give you the same answer. The erroneous answer that you obtained, as I had pointed out, is due to you performing the derivative of s wrongly.

 

[BOAS]

Under which circumstances should I use the quotient rule then?

The quotient rule is useful when you have a problem in the form $f(x) = \frac{u}{v}$ where $u$ and $v$ are functions of x.

For example;

$f(x) = \frac{x^2 + 3x}{2x^2 + 5}$

 

[D H]

Homework Statement

$$\frac{\text{d}}{\text{d}x}(\frac{s}{x})$$
Quotient rule : $\frac{u'v-uv'}{v^2}$
So $\frac{1x-s1}{x^2}$
$=\frac{-s}{x}$
Wolfram says its $\frac{-s}{x^2}$

You used the quotient rule incorrectly. The derivative of a constant is zero. Setting $u(x)=s$ and $v(x)=x$, then $u'(x)=0$ and $v'(x)=1$. Applying the quotient rule correctly yields $\frac {d}{dx} \frac s x = \frac{0*x - 1*s}{x^2} = -\frac s {x^2}$.

 

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\frac{d}{dx} \left(\frac{u}{v}\right) = \frac{du}{dx} \left(\frac{1}{v}\right) + u \frac{d}{dx}\left(\frac{1}{v}\right) = \frac{u'v - uv'}{v^{2}}

One thing I don't understand here: There is no x here,so how can you differentiate with respect to x?

The erroneous answer that you obtained, as I had pointed out, is due to you performing the derivative of s wrongly.

[STRIKE]How did I perform it wrongly?[/STRIKE]
EIDT: Thanks DH, I now know that s is a constant

 

[adjacent]

You used the quotient rule incorrectly. The derivative of a constant is zero. Setting $u(x)=s$ and $v(x)=x$, then $u'(x)=0$ and $v'(x)=1$. Applying the quotient rule correctly yields $\frac {d}{dx} \frac s x = \frac{0*x - 1*s}{x^2} = -\frac s {x^2}$.

What if s is not a function of x but a function of p,which is not related to x in anyway?

 

[D H]

What if s is not a function of x but a function of p,which is not related to x in anyway?

Now you're getting into multivariable calculus. Do you know anything about that subject yet?

If p is unrelated to x whatsoever, your s(p) is still essentially a constant with regard to x. For example, consider $d/dx (p^2/x)$. You have stipulated that p is not related to x in any way. This means $dp/dx=0$, and hence so is $d/dx\,(p^2)$.

 

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Now you're getting into multivariable calculus. Do you know anything about that subject yet?

No. I haven't even studied calculus yet. I am self learning it.

 

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I have another question.
Why is $$\lim_{h \to 0}\left(\frac{0}{h}\right)=0$$?

Shouldn't it be $\frac{0}{0}$

 

[D H]

You should have fully understood limits before you started self-learning calculus.

The concept of the limit of a function f(x) as x→a entails investigating of the behavior of f(x) for values of x near a, but not equal to a. So let's look at the function f(x)=0/x as x→0. This function is identically zero for all values of x except for x=0. The limit never looks at that specific value of x. It instead looks at nearby values of x, and there the function is always zero, so $\lim_{x \to 0} \frac 0 x = 0$.

 

[adjacent]

You should have fully understood limits before you started self-learning calculus.

The concept of the limit of a function f(x) as x→a entails investigating of the behavior of f(x) for values of x near a, but not equal to a. So let's look at the function f(x)=0/x as x→0. This function is identically zero for all values of x except for x=0. The limit never looks at that specific value of x. It instead looks at nearby values of x, and there the function is always zero, so $\lim_{x \to 0} \frac 0 x = 0$.

So how do we calculate limits? How do we know that the function will be 0 when $h \to 0$ without calculating it manually? What if the function is 0.0000000...1 when $h \to 0$? Who knows?

 

[adjacent]

$$\text{D H}$$?

 

[D H]

What is $f(x)=\frac 0 x$ for all x except for x=0?

Have you worked through the epsilon-delta definition of a limit? That same concept is used in the definition of the derivative of a function.

 

[Ray Vickson]

I have another question.
Why is $$\lim_{h \to 0}\left(\frac{0}{h}\right)=0$$?

Shouldn't it be $\frac{0}{0}$

Before trying to learn calculus you should learn---thoroughtly and properly---standard algebra. You ought to know that you can never, ever, under any circumstances, divide by zero. Therefore, you should know that you cannot ever make sense of something like $\frac{0}{0}$.

Anyway, what are limits about? When we say that $\lim_{x \to a} f(x) = L$, what do we mean? We mean (roughly) that as $x$ comes closer and closer to $a$, the values of $f(x)$ come closer and closer to $L$. More precisely, for any pre-assigned 'tolerance' $\epsilon > 0$ we can always force $f(x)$ to lie in the interval $(L-\epsilon,L+\epsilon)$ by making $x$ come close enough to $a$; that is, there is a $\delta > 0$ so that if $a - \delta < x < a + \delta$ then we are guaranteed to have $L-\epsilon < f(x) < L+\epsilon$. (This is known as the so-called epsilon-delta definition, but I hope you realize that $\epsilon$ and $\delta$ are just names; we could equally well talk about the s-t definition, or the u-v definition, or whatever. What matters are the concepts, not the names.

So, in the case of $f(h) = 0/h$, the limit is 0 because as $h$ gets smaller and smaller (in magnitude) 0 gets closer and closer to 0---in fact, it never strays from the value 0.

 

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$h$ gets smaller and smaller (in magnitude) 0 gets closer and closer to 0---in fact, it never strays from the value 0.

How do you know that x gets closer and closer to zero? How can you prove that x is not getting closer and closer to say, $10^{-100000000...}$ which is approximately zero?

 

[Ray Vickson]

How do you know that x gets closer and closer to zero? How can you prove that x is not getting closer and closer to say, $10^{-100000000...}$ which is approximately zero?

By definition: we are asking for a limit as $x \to 0$. Approximately 0 is not the same as 'equal to zero'.

You say you are self-studying, and, frankly, judging from this post and several others, I am worried that you are leading yourself far astray. I don't know what study materials you are using, but they are either not very good or you are mis-using them, or something along those lines.

 

[D H]

that is, there is a $\delta > 0$ so that if $a - \delta < x < a + \delta$ then we are guaranteed to have $L-\epsilon < f(x) < L+\epsilon$.

That's not quite right because $x=a$ satisfies $a-\delta < x < a+\delta$. One of the key points of a limit is that the function f(x) need not be defined at x=a to have a limit at x=a. That singular point x=a must be excluded from the region where f(x) is being evaluated. One way to specify that point from the region of interest is to use $0 < |x-a| < \delta$ .

 

[adjacent]

You say you are self-studying, and, frankly, judging from this post and several others, I am worried that you are leading yourself far astray. I don't know what study materials you are using, but they are either not very good or you are mis-using them, or something along those lines.

No. Everyone learn from mistakes. I am not even learning it too much. I don't have time to do that(Exams all around). I am just preparing myself to learn it in the next holiday(6 months). I will buy some books after this frustrating GCE.