Differentiate the function

1. May 15, 2013

Mary4ever

1. The problem statement, all variables and given/known data
Differentiate:

2. Relevant equations
y=∛((7-3x^2)^2)

3. The attempt at a solution
y^'(x)=(4x(3x^2-7))/(∛((7-3x^2 )^2))^2 )

2. May 15, 2013

Staff: Mentor

It's much more convenient to write your function as y = (7 - 3x2)2/3. When you're differentiating, it's almost always better to rewrite expressions with radicals using exponents. Your answer might be correct, but if so, it needs to be simplified.

3. May 16, 2013

Mary4ever

I really need to know if it is correct, could you please let me know if it is?

4. May 16, 2013

tiny-tim

(try using the X2 button just above the Reply box )
that looks correct, but it needs simplifying

(also, it's a ridiculous way of writing it, and you'll lose marks in the exam if you do that)

5. May 16, 2013

Mary4ever

Thank you for you reply. So what would be the correct way to write it so I do not lose marks for it?

6. May 16, 2013

tiny-tim

first, simplify it, as Mark44 suggested

7. May 16, 2013

Mary4ever

Is this correct: y = (((7-3x^2)^2))^ (1/3)
dy/dx = (1/3) (((7-3x^2)^2))^ (-2/3) (2(7-3x^2)(-6x))
= (24/9)(7x - 3x^3) / ∛((7-3x^2)^4)

8. May 16, 2013

tiny-tim

i'm sorry, but can't you see how ridiculous that is?

why not just write y = (7-3x2)2/3 ?

9. May 16, 2013

Mary4ever

ok so is this a correct final answer now: y'(x)=4x(3x^2-7) ??

10. May 16, 2013

cmcraes

The solution is

12x(3x^(2)-7) isn't it?

11. May 16, 2013

cmcraes

Sorry never mind thought you said derivative not differentiate!

12. May 16, 2013

tiny-tim

how did you get that?

13. May 16, 2013

Staff: Mentor

No.
You get the derivative when you differentiate a function. It seems that you didn't know this.

14. May 16, 2013

Mary4ever

I am really confused now, what is the correct answer then?

15. May 16, 2013

Staff: Mentor

This was the advice I gave in post #2.
What do you get if you differentiate the function that I wrote? You need to use the chain rule (correctly).

16. May 16, 2013

Mary4ever

Differentiating the function you wrote would get: y'(x)=12x^3-28x