1. ## differentiating polynmials

the sum of the two shorter sides of a right-angled triangle is 18cm. calculate the least possible length of the hypotenuse.
ok so obviously i get sqrt(x^2 + (18-x)^2) but how do i differentiate the whole expression from here?

2. $f(x)=\sqrt{(x^2 + (18-x)^2) }$

$f(x)=\sqrt{2x^2-36x+324}$

$f'(x)=\frac{4x-36}{2\sqrt{2x^2-36x+324}}=\frac{2x-18}{\sqrt{2x^2-36x+324}}$

3. Hm... maybe going into more details would be better...

It's like differentiating $(f(x))^n$ where $n \neq 1$

The derivative is:

$n (f(x))^{n-1} \times f'(x)$

$f(x) = \sqrt{x^2 + (18-x)^2} = (2x^2 -36x + 324)^{\frac{1}{2}}$

$f'(x) = \frac{1}{2} (2x^2 -36x + 324)^{-\frac{1}{2}} \times (4x - 36)$

Which you then simplify to:

$f'(x) = \frac{2x - 18}{\sqrt{2x^2 -36x + 324}}$

4. Originally Posted by furor celtica
the sum of the two shorter sides of a right-angled triangle is 18cm. calculate the least possible length of the hypotenuse.
ok so obviously i get sqrt(x^2 + (18-x)^2) but how do i differentiate the whole expression from here?
note that if you minimize $x^2 + (18-x)^2$ , you will also minimize its square root.

5. but how does this fit in with f'(x)=0?

6. 2x^2 -36x + 324 != 0 for all x in R.

Hence sqrt(2x^2 -36x + 324) !=0 for all x in R.

==>

2x-18=0
x=9

7. Wait! If the denominator becomes 0, the f'(x) becomes infinite!

It's the numerator which you should set to zero, not the denominator.