## nickymarden Group Title derive the equation. 2 years ago 2 years ago

1. nickymarden Group Title

2. Rogue Group Title

Senx? What kind of math is this? :o

3. nbouscal Group Title

Is that supposed to be sin x?

4. nickymarden Group Title

sorry, it's in portugese :) and yes it is.

5. Rogue Group Title

Probably. But what is the question asking for? Derive it from...?

6. nbouscal Group Title

Okay, so $$g(x)=(\sin x+\sqrt{x})^3\cdot\sqrt{x^4+2}$$ and you're looking for $$g'(x)$$?

7. lgbasallote Group Title

no wonder i have seen these sen x for a lot of times i thought they taught secret trigonometry to people -___-

8. nbouscal Group Title

So, use the product rule and the chain rule, yeah? Which part are you getting stuck at?

9. nickymarden Group Title

i'm sorry for taking long. I was tryig to do it by myself. I just can't do it lol

10. nbouscal Group Title

Show us your work so we can see where you're getting lost.

11. nbouscal Group Title

First step, product rule, right? So $$f(x)=(\sin x+\sqrt{x})^3, g(x)=\sqrt{x^4+2}$$, and we're finding $$f'(x)g(x)+g'(x)f(x)$$

12. nickymarden Group Title

yeeah

13. Rogue Group Title

$g(x) = (\sin x + \sqrt x)^3 *\sqrt {x^4 +2}$This gets messy...$\frac {dg}{dx} = (\sin x + \sqrt x)^3 \frac {d}{dx} \left[ \sqrt {x^4 +2} \right] + \sqrt {x^4 +2} \frac {d}{dx} \left[ (\sin x + \sqrt x)^3 \right]$

14. Rogue Group Title

As nbouscal said, where do you get stuck?

15. nickymarden Group Title

yeah, i get lost in all the algebra.

16. nickymarden Group Title

i just erased it all, let me try again lol.

17. nbouscal Group Title

Take it one step at a time. First step is to find f'(x), so derive $$(\sin x + \sqrt x)^3$$ using the chain rule.

18. nickymarden Group Title

thank you for the help :))