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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:

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

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

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    $\begingroup$ As the first function is a product, apply the product rule. The second function is a quotient, so apply the quotient rule. BTW: to examine monotony, only the first derivative is needed. $\endgroup$ Nov 18 '18 at 9:58
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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:

$$\frac{\frac{d}{dx}(u).v -u\frac{d}{dx}(v)}{v^2}$$ 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:

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

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  • $\begingroup$ Thanks so much this really help, i was having problem determine the rules to use to solve it but know i understand better $\endgroup$
    – sam
    Nov 18 '18 at 13:21
  • $\begingroup$ @sam I am very happy, you found this useful. Please learn this subject well as it is used in many other topics too. There are plenty of courses and material on the net that give much more details on the subject, such as Khan Academy. Good luck. $\endgroup$
    – NoChance
    Nov 18 '18 at 16:48

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