What is the Derivative as a Limit?

[Karol]

Homework Statement

Homework Equations

Derivative as a limit:
$$y'=\lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$$

The Attempt at a Solution

$$f'(x)=\lim{\Delta x\to 0}\frac{f(x)f(\Delta x)-(1+xg(x))}{\Delta x}=\bigstar$$
$$\left\{ \begin{array}{l} f(\Delta x)=1+\Delta x \cdot g(\Delta x)=1 \\ \Delta x\rightarrow 0 \end{array} \right.$$
$$\bigstar=\lim{\Delta x\to 0}\frac{f(x)\cdot 1-1-xg(x)}{\Delta x}=?$$

 

[fresh_42]

Homework Statement

View attachment 210888

Homework Equations

Derivative as a limit:
$$y'=\lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$$

The Attempt at a Solution

$$f'(x)=\lim{\Delta x\to 0}\frac{f(x)f(\Delta x)-(1+xg(x))}{\Delta x}=\bigstar$$
$$\left\{ \begin{array}{l} f(\Delta x)=1+\Delta x \cdot g(\Delta x)=1 \\ \Delta x\rightarrow 0 \end{array} \right.$$
$$\bigstar=\lim{\Delta x\to 0}\frac{f(x)\cdot 1-1-xg(x)}{\Delta x}=?$$

I don't understand how you get to the formulas behind the brace, but why didn't you go on with this method and substitute the remaining occurrences of $f(x)$ and $f(\Delta x)$ in the formula with the $\bigstar$?

 

[Karol]

$$\bigstar=\lim_{\Delta x\to 0}\frac{1+xg(x)-1-xg(x)}{\Delta x}=\frac{0}{0}$$

 

[fresh_42]

No. You need the $f(\Delta x)$ term as well. Where has it gone to? Looks like an explosion took place and threw it out of the way.

 

[Karol]

But as i showed the $~f(\Delta x)=1$:
$$\lim_{\Delta x \to 0} f(\Delta x)=1+\Delta x \cdot g(\Delta x)=1+0\cdot 1=1$$

 

[Biker]

That is correct, But it is not what we are concerned about. You can't just get the limit of some part of the question and substitute it back. You can however do this.
$\lim_{dx \rightarrow 0} {\frac{f(x)f(dx)}{dx}} + \lim_{dx \rightarrow 0} {\frac{f(x)}{dx}}$
But as you see here it gives none sense. What you need to do is turn the originial equation of the limit
Into something you can get the limit of. As the question noted $\lim_{x \rightarrow 0} {g(x)} = 1$
Remember, The the derivative as a whole has a limit but if you divide the expression it might not.

Consider not substituting $f(x)$ and see what that takes you.

 

[Karol]

$$f'(x)=\lim_{\Delta x\to 0}\frac{f(x)f(\Delta x)-f(x)}{\Delta x}=\frac{f(x)[f(\Delta x)-1]}{\Delta x}=\frac{f(x)[1+\Delta x\cdot g(\Delta x)-1]}{\Delta x}=$$
$$=\frac{f(x)\cdot\Delta x\cdot g(\Delta x)}{\Delta x}$$
$$f'(x)=\lim_{\Delta x\to 0}[f(x)g(\Delta x)]=f(x)\cdot 1$$

 

[Biker]

$$f'(x)=\lim_{\Delta x\to 0}\frac{f(x)f(\Delta x)-f(x)}{\Delta x}=\frac{f(x)[f(\Delta x)-1]}{\Delta x}=\frac{f(x)[1+\Delta x\cdot g(\Delta x)-1]}{\Delta x}=$$
$$=\frac{f(x)\cdot\Delta x\cdot g(\Delta x)}{\Delta x}$$
$$f'(x)=\lim_{\Delta x\to 0}[f(x)g(\Delta x)]=f(x)\cdot 1$$

Excellent so the derivative exists, and it is equal to the function itself

 

[Karol]

Thank you fresh and Biker