Good day! I encountered a problem while doing some differentiation questions and I need some help. The question is:
$\displaystyle f(x)=\frac{x(1-x)(2-x)(3-x)(4-x)(5-x)(6-x)(7-x)(8-x)(9-x)}{(1+x)(2+x)(3+x)(4+x)(5+x)(6+x)(7+x)(8+x)(9+x)} $
Find $f'(0)$.
I know you can just try multiplying everything and using product and quotient rule, but is there a trick to solving this question?
You have $f(x)=xg(x)$, where $g$ is continuous in a neighborhood of $0$; apply the definition: $$ f'(0)=\lim_{x\to0}\frac{xg(x)-0g(0)}{x}=g(0) $$
$f(x)$ has the form $\frac {xg(x)} {h{(x)}}$. So $f'(x)=\frac { h(x)xg'(x)+h(x)g(x)-xg(x)h'(x)} {h(x)^{2}}$. Clearly $f'(0)= \frac {h(0)g(0)} {h(0)^{2}}=1$.
The answer is 1. The good thing here is that when you see the expressions that come out from applying the product rule, most of the terms have a factor $x$ multiplied to them. This makes those terms equal to zero immediately and you will also see that the numerator and denominator on applying the quotient rule and substituting zero turn out to be the same.
Recall that the Taylor series of $\frac{1}{a+x} = \frac{1/a}{1+x/a}$ is:
$$\frac{1}{a} - \frac{x}{a^2} + \frac{x^2}{a^3} - O(x^4)$$
as this is an infinite geometric series with first term $\frac{1}{a}$ and common ratio $-\frac{x}{a}$.
Now $\frac{a-x}{a+x} = \frac{2a}{a+x} -1$. The Taylor series of this is $2a$ times the first series or $\frac{2a}{a} - \frac{2ax}{a^2} + \frac{2ax^2}{a^3} - \cdots$ or $1 - \frac{2x}{a} + \frac{2x^2}{a^2} - \cdots$.
Therefore the Taylor series of $f(x)$, discarding higher-order terms is: $$x \left(1 - \frac{2x}{1} \right) \left(1 - \frac{2x}{2} \right) \left(1 - \frac{2x}{3} \right) \cdots \left(1 - \frac{2x}{10} \right) $$
$$= x \left(1 - \left(\frac{2x}{1} + \frac{2x}{2} + \cdots + \frac{2x}{10} \right) + O(x^2) \right)$$
$$= x - x^2 (\text{some constant} ) + O(x^3)$$
and therefore $f'(x)$ equals:
$$= 1 - 2x (\text{some constant} ) + O(x^2)$$
which when evaluated at $x=0$ gives $1$. Thus $f'(0) = 1$.
Use logarithmic differentiation. $$f(x)= x \frac{\prod_{i=1}^n (i-x) }{\prod_{i=1}^n (i+x) }$$ $$\log(f(x))=\log(x)+\sum_{i=1}^n \log(i-x)-\sum_{i=1}^n \log(i+x)$$ $$\frac{f'(x)}{f(x)}=\frac 1 x-\sum_{i=1}^n \frac 1{i-x}-\sum_{i=1}^n \frac 1{i+x}$$ $$f'(x)=f(x) \times\frac{f'(x)}{f(x)}=\left(\frac 1 x-\sum_{i=1}^n \frac 1{i-x}-\sum_{i=1}^n \frac 1{i+x} \right)x \frac{\prod_{i=1}^n (i-x) }{\prod_{i=1}^n (i+x) }$$ Isolating the first term
$$f'(x)=\frac{\prod_{i=1}^n (i-x) }{\prod_{i=1}^n (i+x) }-\color{red}{x}\frac{\prod_{i=1}^n (i-x) }{\prod_{i=1}^n (i+x) }\left(\sum_{i=1}^n \frac 1{i-x}+\sum_{i=1}^n \frac 1{i+x} \right)$$ Because of the $\color{red}{x}$ in the front of the second term, it is $0$ for $x=0$. The first term is equal to $1$ since for $x=0$ numerator and denominator are identical.
