Find the derivative

Calculus Level 3

$f(x)=x(x-1)(x-2)\cdots (x-9)(x-10)$

Given that $f'(4)=2^a3^b5^c$ for integers $a,b$ and $c$ , find the value of $a+b+c$ .

The answer is 11.

2 solutions

Chan Lye Lee
Nov 14, 2015

Using the definition of derivative (also known as derivative from First Principle), $f'(4)=\lim_{x\to 4}\frac{f(x)-f(4)}{x-4}=\lim_{x\to 4}x(x-1)(x-2)(x-3)(x-5)(x-6)\ldots (x-10)=4!6!=2^73^35^1$

I used product rule and I prefer your solution! Elegant and amazing!

展豪張 - 5 years, 6 months ago

Beautiful and elegant solution!

Jesus Manrique - 5 years, 7 months ago

Exactly how i did it.

Tay Yong Qiang - 5 years, 6 months ago

Stewart Gordon
Nov 12, 2015

By the product rule, $f'(x) = (x-1)(x-2)\cdots (x-9)(x-10)\\ + x(x-2)\cdots (x-9)(x-10)\\ + x(x-1)\cdots (x-9)(x-10)\\ + \cdots \\ + x(x-1)\cdots (x-8)(x-10)\\ + x(x-1)\cdots (x-8)(x-9)$ Observe that each of these terms contains a factor of $(x-4)$ , and therefore equals zero in the evaluation of $f'(4)$ , except for one. Therefore $f'(4) = 4 \times 3 \times 2 \times 1 \times (-1) \times (-2) \times (-3) \times (-4) \times (-5) \times (-6)\\ = \boxed{2^7 3^3 5}$

Thanks for the answer! Alternatively, you may consider $f(x)=(x-4)\times g(x)$ where $g(x)=x(x-1)(x-2)(x-3)(x-5)(x-6)(x-7)(x-8)(x-9)(x-10)$ , and then apply the product rule.

Chan Lye Lee - 5 years, 7 months ago

Same method. Great problem

Michael Fuller - 5 years, 7 months ago

Find the derivative

The answer is 11.

2 solutions

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