How to derive an absolute value function using the chain rule?

Question

I've learned the chain rule in calculus today, but my teacher never explained how to use it with an absolute value function. I think I get the general idea...but could someone check?

Problem: Let f be a real-valued function defined by f(x) = sin^3x + sin^3 x abs(x).

Now: Find f ' (x) for x >0 and f ' (x) for x < 0.

Is it like this when you derive it...?

f ' (x) = 3 sin^2 x + 3 sin^2 x abs(x) x 1

But how does the absolute value work out with the 2 questions...? Please help!

Answer 1

First, your derivatives of your sin functions are wrong. sin^3x is equal to (sinx)^3. Therefore, it needs the chain rule. You started off right, but after you put the 3 out front and decrease the exponent, you also need to multiply by the derivative of the inside function (sinx), which is cosx.

The derivative of absx is actually x/abs(x). This is because abs(x) = the square root of (x^2). This is equal to (x^2)^(1/2). If you find the derivative of this, you get (1/2) multiplied by (x^2)^(-1/2) multiplied by 2x. Simplify and you get x divided by the square root of (x^2), which, as stated, is equal to abs(x).

To summarize: derivative of abs(x) = x/abs(x).

It is the x without the absolute on the top that makes the derivatives of positive and negative numbers different.

Answer 2

You only need to use the chain rule for the sin^3(x). For the second term you need to use the product rule.

Your application of the chain rule was wrong. You did the first part (dy/du) but not the second part (du/dx). You need to take the derivative of sin(x) and multiply it by the derivative of u^3.

If you think about the graph of the absolute value function, it's easy to see what the derivative is, so you can just use the product rule to find the overall derivative of the second term.

By the way, the word is differentiate, not derive. Derive means to find or prove something from something else, not to take the derivative of. Also, when you type a math expression on the computer, use as many parentheses as possible - it makes it much more clear.

Answer 3

Chain Rule Explained

Answer 4

f(x)=(x^(2)−x)^(2/3) f'(x)=(2/3)[(x²-x)^-1/3](2x-1) 0=(2/3)[1/(x²-x)^1/3](2x-1) we find two answers but if we make the graph we find that in the interval (0,1) it does not exist so the minimums are (0,0) and (1,0)