# Math Help - find stationary points

1. ## find stationary points

two problems i have differentiated. need to find stationary point and state if max or min value.

$(a)y= (3x+1)^4$

$dy/dx= 12(3x+1)^3$

this is where im stuck. do i make dy/dx= 0?

$(b)x=\frac{1}{4t^2+3}$

$dx/dt = -8t(4t^2+3)$

many thanks

2. Originally Posted by decoy808

$(a)y= (3x+1)^4$

$dy/dx= 12(3x+1)^3$

this is where im stuck. do i make dy/dx= 0?

Yep...

3. Originally Posted by decoy808

$(b)x=\frac{1}{4t^2+3}$

$dx/dt = -8t(4t^2+3)$
This is not correct

4. Originally Posted by decoy808
two problems i have differentiated. need to find stationary point and state if max or min value.

$(a)y= (3x+1)^4$

$dy/dx= 12(3x+1)^3$ use the chain rule for the power of a function

this is where im stuck. do i make dy/dx= 0?

$(b)x=\frac{1}{4t^2+3}$

$dx/dt = -8t(4t^2+3)$

many thanks

Hi decoy808,

you've erred in your derivative calculation, though it would not matter in calculating the stationary point in this case!

$\frac{d}{dx}\left(3x+1\right)^4=\frac{d}{dx}u^4=\f rac{du}{dx}\frac{d}{du}u^4=4u^3(3)=3\left(3x+1\rig ht)^3$

At a stationary point, this is zero, so $3x+1=0$

$x=\frac{1}{4t^2+3}=\left(4t^2+3\right)^{-1}$

$\frac{dx}{dt}=\frac{dx}{du}\frac{du}{dt}=-\left(4t^2+3\right)^{-2}(8t)=\frac{-8t}{\left(4t^2+3\right)^2}$

This is zero when t=0

You would have gotten the correct answers in spite of your errors! coincidentally

5. so my value part (a) x=-1/3 ?

6. Yep..