Question

How do i find the derrivative of c(x)=30x+50(sqrt(200^2+(600-x)^2)

Answer 1

can u wire it as an equation plz ?

Answer 2

30x+50(200^2+600-x)

Answer 3

btw, squaring something and taking the square root of it is the equivalent to taking the absolute value of it.

Answer 4

Find the derivative of 30x+50|200^2+600-x|

Answer 5

You missed one ")". @asker

Answer 6

\[c(x)=30x+50*\sqrt{200^2+(600-x)^2}\]
Is this the equation you meant?

Answer 7

yes

Answer 8

Then apply the chain rule and go nuts! :D

Answer 9

Remember that you're able to write squareroots of (for example) x as (x)^(1/2).

Answer 10

\[\large c'(x)=\left(30x+50\cdot \sqrt{200^2+(600-x)^2}\right)'\]
\[\large c'(x)=30 (x)'+50 \cdot \left[ \left( 200^2+(600-x)^2 \right)^{1/2 }\right]'\]
\[\large c'(x)=30+50 \cdot \frac{1}{2\sqrt{200^2+(600-x)^2}}\cdot \left(200^2+(600-x)^2\right)'\]

\[\large c'(x)=30+25 \cdot \frac{1}{\sqrt{200^2+(600-x)^2}}\cdot \left[2(600-x)\cdot (600-x)'\right]\]
\[\large c'(x)=30+25 \cdot \frac{1}{\sqrt{200^2+(600-x)^2}}\cdot \left[(1200-2x)\cdot (-1)\right]\]

Go on... (If I made a mistke , you correct it) ... btw this was the hardest one I ever typed with \[\LARGE \LaTeX\]

Answer 11

Haha damn, kudos you Kreshnik!
I should probably learn writing latex fluently as well :(

Answer 12

Just wanna do math and progress though, hehe.

Answer 13

@Noliec haha.. you'll learn latex don't worry ;) ... what does "kudos" mean ? :$

Answer 14

"Kudos (from Greek κῦδος, meaning 'glory') is an English word meaning acclaim or praise for exceptional achievement." ~Wikipedia

Answer 15

haha thank you... take care my friend, see you around. (I have to go).. Great work. (And don't say : I should learn latex ...) - As much as I see , you already know it ! ;) bye

Answer 16

this derriative doesnt touch my original equation when i graph both of them, maybe im doing something wrong

Answer 17

This question is part of a larger minimization question.