Differentiating f(x)=sqrt(x): Power Rule vs Definition

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Homework Statement

Differentiate the function $f(x)=\sqrt{x}$

The Attempt at a Solution

Power rule:
$$\sqrt{x}=x^{\frac{1}{2}}$$
$$\frac{d}{dx}x^{\frac{1}{2}}=\frac{1}{2}x^{-\frac{1}{2}}$$

This is simple.However,I want to solve it using the definition of the derivative
$$\lim_{\Delta x \to 0}=\frac{\sqrt{x+\Delta x}-\sqrt{x}}{\Delta x}$$

I don't know how I should deal with those surds. I can't simply subtract those two sqrts.

 

[CAF123]

Try multiplying numerator and denominator by $\sqrt{x+\Delta x} + \sqrt{x}$.

 

[adjacent]

Try multiplying numerator and denominator by $\sqrt{x+\Delta x} + \sqrt{x}$.

That will get canceled and I will get the same thing again!

 

[CAF123]

That will get canceled and I will get the same thing again!

Well yes, ofcourse since we are effectively multiplying by one. But the idea here is to multiply out the terms on the numerator and denominator. I'll start you off: $$\lim_{\Delta x \rightarrow 0}\, \frac{\sqrt{x+\Delta x} - \sqrt x}{\Delta x} \cdot \frac{\sqrt{x+\Delta x} + \sqrt{x}}{\sqrt{x+\Delta x} + \sqrt{x}} = \lim_{\Delta x \rightarrow 0}\, \frac{x + \Delta x - x}{\Delta x (\sqrt{x+\Delta x}+ \sqrt{x})}\,\, \dots $$

 

[adjacent]

Ok.I get $$\frac{\Delta x}{\Delta \sqrt{x+\Delta x}+ \Delta \sqrt{x}}$$

How do I multiply a constant with it?
I think it should be this: $c . \sqrt{x}=\sqrt{c^2.x}$ (As This is the same as $\sqrt{c^2}.\sqrt{x}=c.\sqrt{x}$

So after applying this rule,I get:
$$\frac{\Delta x}{\sqrt{x.\Delta x +{\Delta x}^2}+\sqrt{x.{\Delta x}^2}}$$
Then what? I can't get anywhere with this.

 

[CAF123]

I am not quite sure what you did there - remember $\Delta x$ is just a number so it obeys the usual distributivity, so $\Delta x ( \sqrt{x + \Delta x} + \sqrt{x} ) = \Delta x \sqrt{x + \Delta x} + \Delta x\sqrt{x}$ if you wanted to multiply out. But notice this is not required - you are left with a $\Delta x$ on the numerator (as you got) and there is a $\Delta x$ on the denominator. So they cancel and what are you left with?

 

[adjacent]

I am not quite sure what you did there - remember $\Delta x$ is just a number so it obeys the usual distributivity, so $\Delta x ( \sqrt{x + \Delta x} + \sqrt{x} ) = \Delta x \sqrt{x + \Delta x} + \Delta x\sqrt{x}$ if you wanted to multiply out. But notice this is not required - you are left with a $\Delta x$ on the numerator (as you got) and there is a $\Delta x$ on the denominator. So they cancel and what are you left with?

Ohhhhhhh.
Yeah.
$$\frac{\Delta x}{\Delta x(\sqrt{x+\Delta x}+\sqrt{x})}=\frac{1}{\sqrt{\Delta x +x}+\sqrt{x}}$$
$$\lim_{\Delta x \to 0}=\frac{1}{\sqrt{x+0}+\sqrt{x}}=\frac{1}{2.\sqrt{x}}$$Thank you sooo much.
On question though,
How did you know that multiplying by $\frac{\sqrt{x+\Delta x} + \sqrt{x}}{\sqrt{x+\Delta x} + \sqrt{x}}$ would make the weird surds go away?
I have never really learned surds in school.So I don't know how it's working.

 

[HallsofIvy]

Ok.I get $$\frac{\Delta x}{\Delta \sqrt{x+\Delta x}+ \Delta \sqrt{x}}$$

You don't want to take the "$\Delta x$" inside the square root. Just cancel the $\Delta x$ outside the square root with the $\Delta x$ in the numerator. That leaves $\frac{1}{\sqrt{x+ \Delta x}+ \sqrt{x}}$

 

[micromass]

How did you know that multiplying by $\frac{\sqrt{x+\Delta x} + \sqrt{x}}{\sqrt{x+\Delta x} + \sqrt{x}}$ would make the weird surds go away?
I have never really learned surds in school.So I don't know how it's working.

It's a trick that you have to see once to know it. We essentially just apply the identity $(a+b)(a-b) = a^2 - b^2$.

Try to find out the derivative of $f(x) = \sqrt[3]{x}$ as an exercise using the identity $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$.

Even better, try to find an identity for $a^n-b^n$ and try to find the derivative of $f(x) = \sqrt[n]{x}$.

 

[CAF123]

How did you know that multiplying by $\frac{\sqrt{x+\Delta x} + \sqrt{x}}{\sqrt{x+\Delta x} + \sqrt{x}}$ would make the weird surds go away?
I have never really learned surds in school.So I don't know how it's working.

I believe it is called 'rationalizing the numerator or denominator' or something like that. For example, if you have the number $a/\sqrt{b}$, then you can rationalize the denominator (i.e make the surd go away) by simply multiplying by $\sqrt{b}$. But if you do this, then you must also multiply the numerator by $\sqrt{b}$ and so in effect the net effect is that you are simply multiplying by one, so the number is unchanged. The result is then that $a/\sqrt{b} = a\sqrt{b}/b$.

In our case, I looked to rationalizing the numerator. $\sqrt{x+\Delta x} + \sqrt{x}$ does the job. But then I must also multiply by the same expression on the denominator too.

 

[adjacent]

I believe it is called 'rationalizing the numerator or denominator' or something like that. For example, if you have the number $a/\sqrt{b}$, then you can rationalize the denominator (i.e make the surd go away) by simply multiplying by $\sqrt{b}$. But if you do this, then you must also multiply the numerator by $\sqrt{b}$ and so in effect the net effect is that you are simply multiplying by one, so the number is unchanged. The result is then that $a/\sqrt{b} = a\sqrt{b}/b$.

In our case, I looked to rationalizing the numerator. $\sqrt{x+\Delta x} + \sqrt{x}$ does the job. But then I must also multiply by the same expression on the denominator too.

Thanks.I think I have a lot to learn in A levels

It's a trick that you have to see once to know it. We essentially just apply the identity $(a+b)(a-b) = a^2 - b^2$.

Try to find out the derivative of $f(x) = \sqrt[3]{x}$ as an exercise using the identity $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$.

Even better, try to find an identity for $a^n-b^n$ and try to find the derivative of $f(x) = \sqrt[n]{x}$.

I will try them for sure.Thanks!

 

[Mark44]

$$\lim_{\Delta x \to 0}=\frac{1}{\sqrt{x+0}+\sqrt{x}}=\frac{1}{2.\sqrt{x}}$$

Don't write $\lim_{Δx \to ...}$ by itself, as you did above and in your first post. That limit thing has to be attached to the function you're taking the limit of. It goes away when you actually take the limit.

Thank you sooo much.
On question though,
How did you know that multiplying by $\frac{\sqrt{x+\Delta x} + \sqrt{x}}{\sqrt{x+\Delta x} + \sqrt{x}}$ would make the weird surds go away?
I have never really learned surds in school.So I don't know how it's working.

A standard trick when you have a sum or difference of radicals in either the numerator or denominator is to multiply by the conjugate (the same two terms, but separated by the opposite sign). The basic idea is that (a + b)(a - b) = a² - b². This can be useful for getting rid of square roots.

 

[Curious3141]

Just for interest's sake, another way to arrive at the answer is to use the binomial/Taylor series, simplify then take the limit. You'll find that only the first order term is significant. But this way is less satisfying because the proof of the series expansion involves calculus in the first place. The elementary method presented here is far better.

 

[adjacent]

Don't write $\lim_{Δx \to ...}$ by itself, as you did above and in your first post. That limit thing has to be attached to the function you're taking the limit of. It goes away when you actually take the limit.

oops.That was a typo mistake.

Just for interest's sake, another way to arrive at the answer is to use the binomial/Taylor series, simplify then take the limit. You'll find that only the first order term is significant. But this way is less satisfying because the proof of the series expansion involves calculus in the first place. The elementary method presented here is far better.

Binomial theorem!
Is it difficult to learn? Can I learn it now?With knowing only differential calculus?

 

[micromass]

Binomial theorem!
Is it difficult to learn? Can I learn it now?With knowing only differential calculus?

The binomial theorem is easy to learn and you can do it right now. It involves expressions of $(a+b)^n$ where $n$ is a positive integer.
However, Curious3141 meant binomial series, which gives expressions of $(a+b)^n$ where $n$ is much more general. You need to know Taylor series for this.