## MTALHAHASSAN2 one year ago Differentiate each of the following function: y=3/(3-x^2)^2

1. MTALHAHASSAN2

do we have to used the quotient rule??

2. Bossvideogamer21

what is the quotient rule?

3. Bossvideogamer21

@MTALHAHASSAN2 ?

4. anonymous

no don't use it if the numerator is a constant

5. anonymous

think of it as $y=3(3-x^2)^{-2}$ and use the power and chain rule

6. MTALHAHASSAN2

wait what you mean that the numerator is a constant

7. MTALHAHASSAN2

@satellite73

8. MTALHAHASSAN2

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9. MTALHAHASSAN2

then is it be like that

10. anonymous

i mean if the numerator is a number has no variable in it

11. anonymous

and no, that answer is wrong @Jhannybean will help

12. anonymous

$\large y=\frac{3}{(3-x^2)^2} \qquad \implies \qquad y=3(3-x^2)^{-2}$\begin{align} y' &= (3\cdot (-2)) \cdot (3-x^2)^{-2-1} \cdot \frac{d}{dx} (3-x^2) \\&=-6(3-x^2)^{-3} \cdot -2x \\&= -12x(3-x^2)^{-3} \\ \ &= \boxed{-\frac{12x}{(3-x^2)^3}}\end{align}

13. MTALHAHASSAN2

@Jhannybean In my answer at the back of the book they don't have a negative but otherwise its all look good

14. anonymous

What does the back of the book say?

15. MTALHAHASSAN2

16. MTALHAHASSAN2

the

17. MTALHAHASSAN2

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18. MTALHAHASSAN2

@Jhannybean why don't you do it like this

19. anonymous

i cant read all of your work and it looks a little confusing.,

20. MTALHAHASSAN2

Ok so I wrote was Y1=3(3-x^2)^-2 + 3(3-x^2)(-2x)

21. MTALHAHASSAN2

@jhannybean

22. anonymous

-6*-2=12

23. MTALHAHASSAN2

What??

24. anonymous

Jhannybean is correct only -6*-2x=12x

25. MTALHAHASSAN2

Wait which processes Is she using

26. anonymous

Jhannybean you have mentioned -6*-2x=-12x which should be 12x

27. MTALHAHASSAN2

WHY DON'T SHE PUT 3 IN THE RIGHT SIDE

28. MTALHAHASSAN2

@surjithayer

29. anonymous

you can put in the right hand side but$\frac{ d }{ dx }\left( af \left( x \right) \right)=a \frac{ d }{ dx }\left( f \left( x \right) \right)$ where a is a constant but it is good if you put in front.

30. MTALHAHASSAN2

You mean 3 is constant so that's it be 0