MTALHAHASSAN2
  • MTALHAHASSAN2
Differentiate each of the following function: y=3/(3-x^2)^2
Mathematics
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

MTALHAHASSAN2
  • MTALHAHASSAN2
do we have to used the quotient rule??
Bossvideogamer21
  • Bossvideogamer21
what is the quotient rule?
Bossvideogamer21
  • Bossvideogamer21

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
no don't use it if the numerator is a constant
anonymous
  • anonymous
think of it as \[y=3(3-x^2)^{-2}\] and use the power and chain rule
MTALHAHASSAN2
  • MTALHAHASSAN2
wait what you mean that the numerator is a constant
MTALHAHASSAN2
  • MTALHAHASSAN2
MTALHAHASSAN2
  • MTALHAHASSAN2
|dw:1444352168711:dw|
MTALHAHASSAN2
  • MTALHAHASSAN2
then is it be like that
anonymous
  • anonymous
i mean if the numerator is a number has no variable in it
anonymous
  • anonymous
and no, that answer is wrong @Jhannybean will help
Jhannybean
  • Jhannybean
\[\large y=\frac{3}{(3-x^2)^2} \qquad \implies \qquad y=3(3-x^2)^{-2}\]\[\begin{align} y' &= (3\cdot (-2)) \cdot (3-x^2)^{-2-1} \cdot \frac{d}{dx} (3-x^2) \\&=-6(3-x^2)^{-3} \cdot -2x \\&= -12x(3-x^2)^{-3} \\ \ &= \boxed{-\frac{12x}{(3-x^2)^3}}\end{align}\]
MTALHAHASSAN2
  • MTALHAHASSAN2
@Jhannybean In my answer at the back of the book they don't have a negative but otherwise its all look good
Jhannybean
  • Jhannybean
What does the back of the book say?
MTALHAHASSAN2
  • MTALHAHASSAN2
they have to same answer as your but you have a negative sign in front of -12x
MTALHAHASSAN2
  • MTALHAHASSAN2
the
MTALHAHASSAN2
  • MTALHAHASSAN2
|dw:1444355117207:dw|
MTALHAHASSAN2
  • MTALHAHASSAN2
@Jhannybean why don't you do it like this
Jhannybean
  • Jhannybean
i cant read all of your work and it looks a little confusing.,
MTALHAHASSAN2
  • MTALHAHASSAN2
Ok so I wrote was Y1=3(3-x^2)^-2 + 3(3-x^2)(-2x)
MTALHAHASSAN2
  • MTALHAHASSAN2
anonymous
  • anonymous
-6*-2=12
MTALHAHASSAN2
  • MTALHAHASSAN2
What??
anonymous
  • anonymous
Jhannybean is correct only -6*-2x=12x
MTALHAHASSAN2
  • MTALHAHASSAN2
Wait which processes Is she using
anonymous
  • anonymous
Jhannybean you have mentioned -6*-2x=-12x which should be 12x
MTALHAHASSAN2
  • MTALHAHASSAN2
WHY DON'T SHE PUT 3 IN THE RIGHT SIDE
MTALHAHASSAN2
  • MTALHAHASSAN2
anonymous
  • 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.
MTALHAHASSAN2
  • MTALHAHASSAN2
You mean 3 is constant so that's it be 0

Looking for something else?

Not the answer you are looking for? Search for more explanations.