Given g(x)=xf(x), show g'(x)=f(x)+xf'(x).

[MooCow]

Problem:
Assume that f has a derivative everywhere. Set g(x)=xf(x). Using the definition of the derivative, show that g has a derivative and that g'(x)=f(x)+xf'(x).

What I know:
I know the definition of the derivative is [f(x+h)-f(x)]/h. I don't know how to plug it in and solve it though. I tried just plugging it straight in like

g'(x)=[x(f(x+h))-xf(x)]/h

I pulled the x out and the rest was [f(x+h)-f(x)]/h (the actual definition) and assumed it was xf'(x). I don't think that's right though.

Any help would be great and appreciated. Thanks in advance:]

 

[John Creighto]

I tried just plugging it straight in like

g'(x)=[x(f(x+h))-xf(x)]/h

What you want is this:

g'(x)=[(x+h)(f(x+h))-xf(x)]/h

 

[dachikid]

What you are asking is called the "chain rule" You should be able to find all kinds of good stuff on google. Essentially, g(x) is defined everywhere, simply meaning that it is continuous (no breaks or sharp point)

To find the derivative of g(x) which is g'(x) you take the first component (i.e. x) and multiply it by the derivative of the second component (i.e. f'(x)). Then you take the second component (i.e. f(x)) and multiply it by the derivative of the first component (i.e. x' = 1). All this should lead you to your solution of g'(x)=f(x) + xf'(x)

Here is a break down step by step
g(x)=xf(x)
Step1
x*f'(x) ==> first component * derivative of the second component

Step 2
f(x) * 1 ==> second component * derivative of the first component

Step 3
Now add those two together
xf'(x)+f(x) to arrive at g'(x)=xf'(x)+f(x)

Yet another way to remember it
first times the derivative of the second + second times the derivative of the first.

Hope this helps!

 

[John Creighto]

What you are asking is called the "chain rule"

I think it would be cheating to use the chain rule in the above problem.

 

[MooCow]

I think it would be cheating to use the chain rule in the above problem.

Yea it would be. Technically I haven't learned that yet so it wouldn't be appropriate to use future processes.

However, even after obtaining the equation you gave me earlier, I still don't know what to do. I don't think I can pull out an (x+h) but that's the only thing I can think of doing...

 

[John Creighto]

Yea it would be. Technically I haven't learned that yet so it wouldn't be appropriate to use future processes.

However, even after obtaining the equation you gave me earlier, I still don't know what to do. I don't think I can pull out an (x+h) but that's the only thing I can think of doing...

When someone mentioned the chain rule above it was a hit.

You don't pull out an (x+h). In one term you pull out an x, the other therm is just f(x). The above problem is a special case of the product rule of differentiation. What you are looking to do is to separate the definition for the derivative of f(x) from the rest of the terms in your expression. You can separate these terms via product or sum.

 

[Rainbow Child]

In the numerator of
$$g'(x)=\lim_{h\to 0}\frac{(x+h)\,f(x+h)-xf(x)}{h}$$

add and subtract the term $(x+h)\,f(x)$ and try to form the term

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

 

[John Creighto]

In the numerator of
$$g'(x)=\lim_{h\to 0}\frac{(x+h)\,f(x+h)-xf(x)}{h}$$

add and subtract the term $(x+h)\,f(x)$ and try to form the term

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

I don't know. That might work. There is a more obvious way to solve it though.

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

The rest is left as an exercise.

 

[Rainbow Child]

It does work, because that's the way you prove the chain rule

 

[John Creighto]

It does work, because that's the way you prove the chain rule

I thought that was the case but it is certainly not the most obvious method.