# Finding derivative and solve for t

Hello so I'm supposed to find the derivative of (-t2+4)/(t2+4)2

and using the quotient rule + the chain rule I got this mess:
dy/dx= -2t(t^2+4)^2-2t(2t2+8)(t2+4)2/ (t2+4)4

and now I have to set that whole thing to 0. I know the numerator must be equal to 0 which mean that one of the solution is t=0. Is there any other solutions?Can you help me simplify/factor it to find the other solution?Thank you

Dick
Homework Helper
Your 'mess' is a ways from being right yet. Once you get it right I'd start with canceling out the common factors of (t^2+4) in the numerator and denominator. It will look a lot less messy.

Mark44
Mentor
josh_123 said:
Hello so I'm supposed to find the derivative of (-t2+4)/(t2+4)2
and using the quotient rule + the chain rule I got this mess:
dy/dx= -2t(t^2+4)^2-2t(2t2+8)(t2+4)2/ (t2+4)4
and now I have to set that whole thing to 0. I know the numerator must be equal to 0 which mean that one of the solution is t=0. Is there any other solutions?Can you help me simplify/factor it to find the other solution?Thank you
Since there is no x in your original expression, your derivative shouldn't be dy/dx. Also, there's no y in that expression, either, so where did it come from?

Why do you think you need to set the derivative to 0? The question merely asks you to find the derivative.

Hello there,
the function is v(t)=-t2+4/(t2+4)2
and to find the derivative I'm suppose to find v'(t) or a(t)
The reason why I have to set the second derivative to 0 so that I could find the point at which the function is concave up or down. (those are called stationary points?)
I find the second derivative by using the equation:
f'g-g'f/g^2
and there is a comment said that my second derivative is wrong. Can you guide me where I make the mistake?
Thank you

It could have gone wrong anywhere. Show us your steps in simplifying your derivative after applying the quotient rule. Only then will we be able to pinpoint your mistake. So when you set your 2nd derivative = 0, you are looking at when there is constant velocity or 0 acceleration. I wouldn't call these stationary points since the object is still moving in some direction. Stationary points would only be when v(t)=0

oh okay here is the step
a(t)= -2t(t^2+4)^2 is the part f'g and then subtract the g'f part I used the chain rule to find g' which is 2t(2t^2+8)<----g'(t^2+4)^2<-----f and divide by g^2 which is( (t^2+4)^2)^2) which clean up to be equal (t^2+4)^4

and that why I got -2t(t^2+4)^2-2t(2t^2+8)(t^2+4)^2/ (t^2+4)^4

Dick
Homework Helper
oh okay here is the step
a(t)= -2t(t^2+4)^2 is the part f'g and then subtract the g'f part I used the chain rule to find g' which is 2t(2t^2+8)<----g'(t^2+4)^2<-----f and divide by g^2 which is( (t^2+4)^2)^2) which clean up to be equal (t^2+4)^4

and that why I got -2t(t^2+4)^2-2t(2t^2+8)(t^2+4)^2/ (t^2+4)^4

The f'g part looks ok. The g'f part doesn't. It looks more like g'g to me.

SammyS
Staff Emeritus
Homework Helper
Gold Member
(You really do need to learn to use parentheses where they're needed.)

I take it that you have f(t)/g(t) = (-t2+4)/(t2+4)2 .

Then $\displaystyle \frac{d}{dt}\left(\frac{f(t)}{g(t)}\right)=\frac{f\,'\!(t)\,g(t)-f(t)\,g'(t)}{(g(t))^2}$

I don't see f(t) in the second term of the numerator in your expression.

Mark44
Mentor
the function is v(t)=-t2+4/(t2+4)2
and to find the derivative I'm suppose to find v'(t) or a(t)
The reason why I have to set the second derivative to 0 so that I could find the point at which the function is concave up or down. (those are called stationary points?)
This information is part of the problem statement, which should have been in your first post.

Hello everyone, so I redo the derivative

a(t)= -2t(t2+4)2-2(t2+4)(2t)(-t2+4)/ (t2+4)4

Mark44
Mentor
Hello everyone, so I redo the derivative

a(t)= -2t(t2+4)2-2(t2+4)(2t)(-t2+4)/ (t2+4)4

$$a(t) = -2t(t^2 + 4)^2 - \frac{2(t^2 + 4)(2t)(-t^2 + 4)}{(t^2 + 4)^4}$$