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thanks buddy your awesome

righttt pi [(1/-1)x^-1]

(-1/3)x^-3

since it's divided by 1, isn't it negative n =x^-2

\pi [ \frac{x^{-3}}{-3}]

\pi \int^{2}_{1} \frac{1}{x^2} dx

\int^{1}_{2} \pi [\frac{1}{x}]^2 dx if you help me with this problem, i am sure i will get the rest of the problems.

[SOLVED] volumes... last week of calc i was absent when we went over the volume section. y= 1/x, x= 1, x= 2, y= 0,; about the x-axis \int^{1}_{2} \pi \frac{1}{x} dx i don't know what's next. can anyone inform me, please

am soo stupid...thanks

2x^2 - x^2 - 1 = x^2 - 1 it might sound childish, but how is that equal? you can't factor anything or simplify it

\frac{2x^2- (x^2+ 1)}{x^2}= \frac{x^2-1}{x^2} how are they equall?

so, i was right?

how are they equall?

\frac{(2x)- (x^2+1)}{x^2}= \frac{x^2- 1}{x^2} how are those equal?

whaaat

thats what i got to so far (x^2 + 1)/ x 0=({{2x^2}-{x^2}+{1}})/{x^2} which is undifined... NO answer

do u want me to use the quo rule for 1/x which is -1/x^2?

1 + ln x =0 -1 = ln x thats about all i can do

[SOLVED] one last optimization problems Homework Statement find two positive numbers such that the sum of the number and its reciprocal is as small as possible. Homework Equations x+(1/x) = s f' =1 + ln x The Attempt at a Solution lost

thanks alot. care for one more problem?

y= 25

find the derivates of 50x-x^2 50-2x=0 x=25 100=2(25)+2y

y = (100/2)-x

am trying to solve for y

2y = 100/2x y = (100/2x) * .5

Homework Statement find the dimensions of a rectangle with perimeter 100m whose area is as large as possible Homework Equations area = XY 100 = 2x + 2y y= 100/4x x(100/4x) (400x - 400x)/16x^2 1/16x^2 = 0 The Attempt at a Solution well... am lost

thanks alot. am messed up today!

x=\sqrt{1000}=\sqrt{10\cdot10^2}=10\sqrt{10} how did u get that?

x=square root of .001?

2/1000=x^-2

2-(2000/x^2)=0 i think

[SOLVED] simple optimization problem Homework Statement find the dimensions of a rectangle with area 1000 m^2 whose perimeter is as small as possible Homework Equations perimeter = 2x + 2y area = xy 1000 = xy y= 1000/x perimeter = 2x + 2(1000/x) The Attempt at a...

but the b's (x and y) doesn't cancel since they have two different signs

\frac{5x-5y-bx+by}{5x-5y+bx-by} i can't cancel anything i can't factor anything please help

\frac{dy}{dx} {y= 2x^{sinx}} product rule f= 2 f'= 0 g= x g'=? chain rule f= x f'= 1 g=? g'=? is the anser y'=-2x^{cosx}

or maybe the answer is y=-2cosx

y=2x^{sinx} i know i should use the product rule within a chain rule. but how can i use chain rule with sinx is the anwser y=-2x^{cosx} can anyone give me pointer to this easy problem and tell if am forgetting something.

Y'= e^{-x2}(1-2x^{2}) thanks for the help

Solved it thanks

[SOLVED] another chain rule: easy one y=xe^{-x^2} i have no i dea how to start. f'= x^{x^2} or -2x^blah blah blah just get me started and i'll promise you i will finish it myself

g'= -m Solved. thanks for the help

lets just go back to the original equation fsingle-quote(x)=e^{g(x)}gsingle-quote(x) what is g'(X) equall to

never mind, sorry.

ysingle-quote=a+x\cdot0=a times 0 would be zero,right?

gosh! well i would use the product rule or maybe not. f= x g=ax f'= 1 g'=product rule high school calc.

how did you get g'(x) i thought the (m) was constant thus making the the derivative equal to 0

yes!

[SOLVED] first derivative: chain rule: easy for you guys Y=E^(-mx) f= E^x g= -mx f'= E^x g'= o E^(-mx) * 0(E^(-mx)) i think, not sure though Y'= 0 which is wrong someone help

how should i break it down?

Homework Statement A high voltage power line carries a current of 110 A at a location where the Earth's magnetic field has a magnitude of .59 G and points to the north, 72 degrees below the horizontal. find the direction and magnitude of the magnetic force exerted on a 250-m lengh of wire if...