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

• MooCow
In summary, using the definition of the derivative, we can show that g(x) has a derivative and that g'(x)=f(x)+xf'(x). This can be done by separating the terms in the numerator of the derivative expression and using the limit definition of the derivative. Alternatively, we can use the product rule of differentiation to arrive at the same solution.
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:]

MooCow said:
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

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
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!

dachikid said:
What you are asking is called the "chain rule"

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

John Creighto said:
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...

MooCow said:
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.

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

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

Rainbow Child said:
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.

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

Rainbow Child said:
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.

## 1. What does the notation g(x) = xf(x) mean?

The notation g(x) = xf(x) means that the function g(x) is equal to the product of x and another function f(x). This means that the output of g(x) will be the input x multiplied by the output of f(x).

## 2. How do you find the derivative of a function with this notation?

To find the derivative of a function with this notation, we can use the product rule, which states that the derivative of f(x) multiplied by g(x) is equal to f(x) multiplied by the derivative of g(x) plus g(x) multiplied by the derivative of f(x). In this case, the derivative of g(x) would be equal to f(x) plus x multiplied by the derivative of f(x).

## 3. How does the given equation demonstrate the product rule?

The given equation, g'(x) = f(x) + x*f'(x), demonstrates the product rule because it shows the derivative of a product of two functions, g(x) = xf(x), being equal to the sum of the first function, f(x), multiplied by the derivative of the second function, f'(x), plus the second function, x, multiplied by the derivative of the first function, f'(x).

## 4. Can the product rule be applied to any two functions?

Yes, the product rule can be applied to any two functions, as long as they are differentiable. This means that they must have a defined derivative at every point in their domain.

## 5. How can the product rule be used to find the derivative of a more complex function?

The product rule can be used to find the derivative of a more complex function by breaking it down into simpler functions that can be differentiated using the product rule. The derivative of each simpler function can then be found and combined using the sum rule to find the derivative of the original complex function.

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