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find the derivative $\displaystyle{{{\rm d} \over {\rm d}x}\left[% {\left(x^{3} + 2\right)\left(x^{2} + 2\right) \over x^3+1}\right]}$

This is what I have so far:
$$(x^3+1)[(x^3+2)(x^2+2)]-[(x^3+1)](x^3+2)(x^2+2)/(x^3+1)^2$$ $$(x^3+1)[(3x^2)(2x)]-[3x^2](x^3+2)(x^2+2)/(x^3+1)^2$$

I have the answer but I don't know how to get there: $$2x^7+3x^4-6x^+4x/(x^3+1)^2$$

Gigili
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envy
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4 Answers4

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I guess you may misunderstand how to get $$\{(x^3+2)(x^2+2)\}^\prime.$$

You can use $$g^\prime(x)=(x^3+2)^\prime(x^2+2)+(x^3+2)(x^2+2)^\prime.$$

Let $$f(x)=x^3+1,g(x)=(x^3+2)(x^2+2).$$ First find $f^\prime(x),g^\prime(x)$, then set them in $$\left\{\frac{g(x)}{f(x)}\right\}^\prime=\frac{g^\prime(x)f(x)-g(x)f^\prime(x)}{\{f(x)\}^2}.$$

Note that in general, $$\{h(x)i(x)\}^\prime=h^\prime(x)i(x)+h(x)i^\prime(x).$$

EDIT : Since $$g^\prime(x)=3x^2(x^2+2)+(x^3+2)\cdot 2x=5x^4+6x^2+4x,$$ yours will be $$\frac{(5x^4+6x^2+4x)(x^3+1)-(x^3+2)(x^2+2)\cdot 3x^2}{(x^3+1)^2}.$$

Then, simplify the numerator.

mathlove
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you are complicating things,please multiplying together in brackets and then use derivative rule for quotient polynomials,if you multiply in brackets,you get

$x^5+2*x^3+2*x^2+4$

so finally you have this quotient polynomials

$(x^5+2*x^3+2*x^2+4)/(x^3+1)$

now please consider that $d(f/g)/dx=(f'*g-g'*f)/(g^2)$

http://www.math.hmc.edu/calculus/tutorials/quotient_rule/

EDITED:

ok suppose that we are trying to derivative following one

$(x^3+4)(x^2+x+1)$

now you can use product rule and make derivative,but is not better to multiply this two polynomial on each other and then make derivative? what do you think?

dato datuashvili
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One might also notice that $$ \frac{x^3+2}{x^3+1}=1+\frac1{x^3+1} $$ to simplify before using the quotient rule: $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\frac{(x^3+2)(x^2+2)}{x^3+1} &=\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2+2+\frac{x^2+2}{x^3+1}\right)\\ &=2x+\frac{2x(x^3+1)-3x^2(x^2+2)}{(x^3+1)^2}\\ &=2x+\frac{2x-6x^2-x^4}{(x^3+1)^2} \end{align} $$

robjohn
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d/dx((x³+2)(x²+2)/(x³+1)) = d/dx((x⁵+2x³+2x²+4)/(x³+1))
=((x³+1)(x⁵+2x³+2x²+4)′-(x⁵+2x³+2x²+4)(x³+1)′)/(x³+1)²
=((x³+1)(5x⁴+6x²+4x)-(x⁵+2x³+2x²+4)(3x²))/(x³+1)²
=(5x⁷+6x⁵+4x⁴+5x⁴+6x²+4x-3x⁷-6x⁵-6x⁴-12x²)/(x³+1)²
=(2x⁷+3x⁴-6x²+4x)/(x³+1)²
Now, factor the top, just to be sure nothing cancels out:
=x(2x⁶+3x³-6x+4)/(x³+1)². Nothing can cancel, so you can leave it as you had it before:
=(2x⁷+3x⁴-6x²+4x)/(x³+1)²
MathBob
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