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Hello everyone how would I find derivative $f'(-2)$ if

$$f(x)=(x^3-5)(x^2-3)(x+1)$$

What I did is I made $(x^2-3)(x+1)$ the other part in the product rule.

This be what I did.

$(x^3-5)(2x^2+2x-3)+(3x^2)(x^3+x^2-3x-3)$

Then I plug I in $x=-2$ and got $-25$ but I am not sure if I did it correctly.

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It should be $3x^2+2x-3$ for the second factor of your first term. –  1015 Feb 9 '13 at 22:55
    
Mistake in differentiating $x^3+x^2-3x-3$. –  Gerry Myerson Feb 9 '13 at 22:55
1  
I think the second factor in the first term should have an $3 x^2$, not a $2 x^2$. –  Ron Gordon Feb 9 '13 at 22:55
    
Oh I see let me see If I get a correct answer. –  Fernando Martinez Feb 9 '13 at 23:05

3 Answers 3

up vote 3 down vote accepted

Your strategy works fine here, since $(x^2 - 3)(x+1)$ is fairly easy to expand:

$$f(x) = (x^3 - 5)[(x^2-3)(x+1)]=(x^3 - 5)(x^3 + x^2 - 3x -3)$$

Using the product rule:

$$f'(x) = (x^3-5)(\color{blue}{\bf{3}}x^2+2x-3)+(3x^2)(x^3+x^2-3x-3)$$

(i.e., your only error was in miscalculating the derivative of $x^3$..

Now, evaluate $f'(-2)$. And if you'd like to check your work, feel free to hover the calculation below.

$$f'(-2) = -77$$

Note: The product rule for two factors can be expanded to a product rule for three factors, which is handy, to use with more complicated polynomial factors, and/or other functions of $x$: If you have a function that is a product of three functions, e.g.

$$f(x)=g(x)\cdot h(x)\cdot j(x)$$

Then can apply use the following (Product Rule, three factors):

$$f'(x) = \left(g'(x)\cdot h(x)\cdot j(x)\right) + \left(g(x) \cdot h'(x)\cdot j(x)\right) + \left(g(x)\cdot h(x)\cdot j'(x)\right)$$

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thanks I got $-77$ now. –  Fernando Martinez Feb 9 '13 at 23:09
    
Yup...that's what you want! Your only error was in the mis-calculation of $(x^3)\,'$ –  amWhy Feb 9 '13 at 23:10

Here is a suggestion if you have a longer product. The calculation will not appear to make full sense, since we will be taking the logarithm of things that may be negative. But everything in fact works out fine.

We use $n=4$, but the method is general. suppose we want the derivative of $y$, where $y=f_1(x)f_2(x)f_3(x)f_4(x).$

Taking the logarithm, we find (?) that $$\log y=\log f_1+\log f_2+\log f_3+\log f_4$$ Thus $$\frac{y'}{y}=\frac{f_1'(x)}{f_1(x)}+ \frac{f_2'(x)}{f_2(x)}+\frac{f_3'(x)}{f_3(x)}+\frac{f_4'(x)}{f_4(x)}.$$ Multiply through by $y$, that is, by $f_1(x)f_2(x)f_3(x)f_4(x)$ to find $y'$.

With equal ease, the procedure handles quotients whose numerators and denominators are products.

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The answer I believe should be -77, to check your work.

I'm going to expand a bit on what it is you are doing.

A completely valid way is to multiply out all the terms and take the derivatives that way.

A faster way is using the product rule.

If you have multiple products of functions:

$$f=g\cdot h\cdot j$$

Which in your case you do:

$$f(x)=(x^3−5)(x^2−3)(x+1)$$ $$g=(x^3-5)$$ $$h=(x^2-3)$$ $$j=(x+1)$$ With Derivatives $$g'=3x^2$$ $$h'=2x$$ $$j'=1$$

You apply the following technique (Product Rule) $$f'=g'\cdot h\cdot j + g \cdot h'\cdot j+g\cdot h\cdot j'$$

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