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I have

$\frac{\partial (f(x) g(x))}{\partial x}$=$g(x) f'(x)+f(x) g'(x)$,

I need to differentate this function with respect to x.

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

However I do not see the logic using the product rule.

ALEXANDER
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  • Why not? $f(x)$ in the question is the product of several functions. – Quang Hoang Jul 28 '14 at 03:56
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    $f$ is a product of two function $g(x)=(x+1)(x+2)^2$ and $h(x)=(x+3)^3(x+4)^4$ that are themselves products. With method, this is not difficult to apply. Since it is difficult to be good in Calculus without raising one's sleeves and compute, you should give it a try, and we will correct it.

    Alternatively, it is not very difficult to show that for a product of $3$ functions $(fgh)'=f'gh+fg'h+fgh'$ or for a product of $4$ functions $(fghk)'=f'ghk+fg'hk+fgh'k+fghk'$.

     – Taladris Jul 28 '14 at 04:04
  • Note also that for a function of one variable, the notation $\frac{\partial f}{\partial x}$ is considered incorrect and is reserved to function of several variables. Usual notations are $\frac{d f}{d x}$ or just $f'(x)$. I would also avoid to mix notations for the same thing and write $(fg)'=f'g+fg'$ (or the equivalent with the $d$-notation). – Taladris Jul 28 '14 at 04:08
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    How do you compute $2 \cdot 3 \cdot 4 \cdot 5$ if the only thing you know is how to multiply two numbers? –  Jul 28 '14 at 04:17

3 Answers3

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For this example, it will be useful to use logarithmic differentiation. Let $y$ be our function. Then $$\ln y=\ln(x+1)+2\ln(x+2)+3\ln(x+3)+4\ln(x+4).$$ Differentiate. We get $$\frac{1}{y}\frac{dy}{dx}=\frac{1}{x+1}+\frac{2}{x+2}+\frac{3}{x+3}+\frac{4}{x+4}.$$

André Nicolas
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The product rule states that: $(f_1(x)f_2(x))' = f_1'(x)f_2(x) + f_1(x)f_2'(x)$. This works in general for $n$ functions: $$(f_1(x) \cdots f_n(x))' = f_1'(x)f_2(x)\cdots f_n(x) + f_1(x)f_2'(x)\cdots f_n(x) + \cdots + f_1(x)\cdots f_n'(x)$$ The prime $'$ runs through all of the functions. Writing in a more compact way, we can say that: $$\left(\prod_{i=1}^n f_i(x)\right)' = \sum_{i = 1}^n \left(f_i'(x) \prod_{j = 1, j \neq i}^n f_j(x)\right)$$ You can prove this easily by induction on $n$. Now, applying this on your specific problem, we have: $$f'(x) = (x+1)'(x+2)^2(x+3)^3(x+4)^4 + (x+1)\left((x+2)^2\right)'(x+3)^3(x+4)^4 + \cdots+(x+1)(x+2)^2(x+3)^3\left((x+4)^4\right)'$$ In each factor, you use the chain rule.

Ivo Terek
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    It seems worth noting that the familiar power rule is itself an instance of this, e.g. $$\dfrac{d}{dx}x^3 = \dfrac{dx}{dx}\cdot x \cdot x+x\cdot\dfrac{dx}{dx}\cdot x+ x \cdot x \cdot \dfrac{dx}{dx} =3x^2$$ (at least for natural powers) – Semiclassical Jul 28 '14 at 04:17
  • Nice comment. We can also prove that $(x^n)' = nx^{n-1}$ by using induction and the product rule for two functions. – Ivo Terek Jul 28 '14 at 04:19
  • Do not see what you mean by "in each factor you use the chain rule" – ALEXANDER Jul 28 '14 at 04:47
  • @ALEXANDER for example, doing $$\left((x+3)^3\right)' = 3(x+3)^2 \cdot 1$$ instead of writing everything, like: $$(x+3)^3 = x^3 + 3x^2(3) + 3x(3^2) + 3^3$$ and then differentiating it. – Ivo Terek Jul 28 '14 at 04:54
  • Okey, perfect I see! Thank you! – ALEXANDER Jul 28 '14 at 04:55
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HINT

You must do that step by step.

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

$$f'(x)=((x+1))' ((x+2)^2 (x+3)^3 (x+4)^4)+((x+1)) ((x+2)^2 (x+3)^3 (x+4)^4)'$$

$$((x+2)^2 (x+3)^3 (x+4)^4)'=((x+2)^2)' ((x+3)^3(x+4)^4)+((x+2)^2) ((x+3)^3(x+4)^4)'$$

$$((x+3)^3(x+4)^4)'=((x+3)^3)'((x+4)^4)+((x+3)^3)((x+4)^4)'$$

now you go from the last to the first.

Shabbeh
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