(v^2)(6/√12^v-1)+(√12^v-1)(2v)
S summer.johnson New member Joined Feb 28, 2012 Messages 2 Feb 28, 2012 #1 (v^2)(6/√12^v-1)+(√12^v-1)(2v)
M medicalphysicsguy New member Joined Jan 23, 2012 Messages 28 Feb 29, 2012 #2 That is hard to read. I cannot tell what your square roots apply to. Is this a calculus question or do you just need to factor out some variables? mpg
That is hard to read. I cannot tell what your square roots apply to. Is this a calculus question or do you just need to factor out some variables? mpg
D Deleted member 4993 Guest Feb 29, 2012 #3 summer.johnson said: (v^2)(6/√12^v-1)+(√12^v-1)(2v) Click to expand... Are trying to find \(\displaystyle \frac{df(v)}{dv}\)? As you wrote it - your function looks like: \(\displaystyle f(v) \ = \ (v^2) * (\frac{6}{\sqrt{12^v}} \ - \ 1) \ + \ (\sqrt{12^v} \ - \ 1) * (2*v)\) Is that correct? If not - using proper grouping symbols (and paying attention to order of operations aka PEMDAS) repost your problem.
summer.johnson said: (v^2)(6/√12^v-1)+(√12^v-1)(2v) Click to expand... Are trying to find \(\displaystyle \frac{df(v)}{dv}\)? As you wrote it - your function looks like: \(\displaystyle f(v) \ = \ (v^2) * (\frac{6}{\sqrt{12^v}} \ - \ 1) \ + \ (\sqrt{12^v} \ - \ 1) * (2*v)\) Is that correct? If not - using proper grouping symbols (and paying attention to order of operations aka PEMDAS) repost your problem.
