Question

How do you take the derivative of g(x)=l(3x-4)l

Answer 1

Well, you don't for x = 4/3.

For x > 4/3, the absolute values do nothing. Just discard them
For x < 4/3, what? I'm not telling.

Answer 2

?

Answer 3

You'll have to do better than that. What part are you questioning?

Answer 4

what are you talking about?

Answer 5

What are you reading? It is a direct response to your question. Nothing tricky or evasive.

Answer 6

I don't understand what your trying to say I'm sorry I'm not smart at calculus

Answer 7

Think about how the slope changes for l(3x-4)l as compared to 3x-4

Answer 8

What part of "You don't for x = 4/3." do you not understand?

Answer 9

it would always be positive, because when you take the absolute value the value can never be negitive

Answer 10

Can you graph the slope?

Answer 11

Wouldn't the slope by 4/3

Answer 12

\(g(x) = |3x-4| = \begin{cases}
3x-4\;\;\;\;\;\;\;\text{if}\ x \ge 4/3 \\
-(3x-4)\;\text{if}\ x < 4/3
\end{cases}\)

Answer 13

\(\color{#000000 }{ \displaystyle |z|=\sqrt{z^2} }\)

So,

\(\color{#000000 }{ \displaystyle \frac{dy}{dx}|y|= \frac{d}{dx}\sqrt{y^2}=\frac{1}{2\sqrt{y^2}}\times{\color{blue}{2}y^{2-\color{red}{1}}} \times y' }\)

\(\color{#000000 }{ \displaystyle \frac{dy}{dx}|y|= \frac{d}{dx}\sqrt{y^2}=\frac{2y\cdot y'}{2\sqrt{y^2}} =\frac{y}{|y|}\cdot y' }\)

Answer 14

Just, by applying the definition of the absolute value (my first equation), I can conclude that,

\(\color{#000000 }{ \displaystyle \frac{d}{dx}\left|f(x)\right| =\frac{f(x)\times f'(x)}{\left|f(x)\right|} }\)

Answer 15

In that case, you still don't for x = 4/3.

Answer 16

You have to split the slope into two pieces because g(x) is not continuous.

Answer 17

Yes, at the "vertex" (or whatever the corner is called), the function is not differentiable. try to draw the tangent line through that point (xD) :)

Answer 18

okay that makes sense, not that I think about it because at lxl you can't differentiate at (0,0)

Answer 19

Yes, |x| you can't differentiate at x=0.
You don't have an instantaneous slope there!

However,

You could get an explicit derivative-function
for \(\small\color{#000000 }{ \displaystyle f(x)=\left|mx+b\right| }\), and note that for the derivative,
you have a restriction, and that is: \(\color{#000000 }{ x\ne\frac{-b}{m} }\).

But, for other than this, you just have,

\(\color{#000000 }{ \displaystyle f'(x)=\frac{mx+b}{\left|mx+b\right|} \times m}\)

Answer 20

okay

Answer 21

is that right?