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I'm having a problem finding the first and second derivatives of the following functions so that I can use them to find the monotonic intervals and the curve.

For the first function, I tried to use the chain rule combined with the product rule, but I am not sure I am doing the right thing. For the second function, I don't have any idea how to get the derivatives, because it's in fraction form.

1. $y=(x-1) (x+1)^3$


2. $y=dfrac{10}{4x^3-9x^2+6x}$

For the first function, I tried this:

$$begin{align}
y &= (x-1) (x+1)^3 \
&= (x-1)^{-1}cdot 1cdot 3(x+1)^2 =0
end{align}$$

I get lost from here. Any help on this? Thanks.

I'm having a problem finding the first and second derivatives of the following...

Hint: I will show the first derivative of each function below, hope this helps:

for $y=(x-1) (x+1)^3$

you could use the chain rule: $frac{d}{dx}(u.v)=v.frac{d}{dx}(u)+u.frac{d}{dx} (v)$

Choose $$u=x-1, v=(x+1)^3$$

$$frac{d}{dx}(u)=frac{d}{dx}(x-1)=1$$
$$frac{d}{dx}(v)=frac{d}{dx}(x+1)^3=3(x+1)^{3-1}(x-1)=3(x+1)^2(x-1)$$
$$frac{d}{dx} (x-1) (x+1)^3 =(x+1)^3+(x-1)(3(x+1)^2)$$

Now for the function:
$y=dfrac{10}{4x^3-9x^2+6x}$ This function is of the form $g=frac{u}{v}$ and the derivative is:

$$frac{frac{d}{dx}(u).v -ufrac{d}{dx}(v)}{v^2}$$
choose $$u=10, v= 4x^3-9x^2+6x$$
$$frac{d}{dx}(u)=frac{d}{dx}(10)=0$$
$$frac{d}{dx}(v)=frac{d}{dx} (4x^3-9x^2+6x)=12x^2-18x+6$$

$$frac{d}{dx}(frac{10}{4x^3-9x^2+6x})=frac{-10(12x^2-18x+6)}{(4x^3-9x^2+6x)^{2}}$$

The above derivative is not defined where v=0, that is at values:

$$x=0 , x=frac{9}{8}+ifrac{sqrt {15}}{8}, frac{9}{8}-ifrac{sqrt {15}}{8}$$