differentiation

a) Just consider the equation as y = (1/2)(1/x). Now, you can see that the 1/2 is a constant, and so can be left alone. The 1/x can be written as x^-1. So,

dy/dx = (1/2)(-1)x^-2 = -1/(2x^2)

b) Do the same thing, considering the equation as y = (3/2)(1/x^2) = (3/2)(x^-2)

dy/dx = (3/2)(-2)x^-3 = -3/x^3

Reply 3

username9816

franks

a) y= 1 / 2x
b) y= 3 / (2x^2)

a.) y = 1 / 2x = 1/2 * 1/x = 1/2 * (x^-1)
dy/dx = 1/2 * (-x^-2) = 1/2 * (-1/x^2)
---> dy/dx = -1/(2x^2)

b.) y = 3/(2x^2) = 3/2 * (1/x^2) = 3/2 * (x^-2)
dy/dx = 3/2 * (-2x^-3) = 3/2 * (-2/x^3)
---> dy/dx = -3/(x^3)

Reply 4

OP

okkkk
i can see what you are all doing but i just wanna know why :biggrin:

u know normally when u have y= x^2 + 4 dy/dx = 2x and u ignore the 4 cos theres no x's ... well how come u dont ignore the 1/2 in question 1 or the 3/2 in the 2nd question? - is it cos its multiplied to the x bit..... tell me if that doesnt make sense and i'll try again !

Reply 5

m:)ckel

yeah, you need to keep it because it's the coefficient of the x term (the bit multiplying the x)....just think if you had y = 3x, what would you do to differentiate it?
When you just have a constant on its own, then when differentiated, it becomes 0

Reply 6

OP

ok, thanks a lot... thats cleared that one up :smile:

i'll go and see if i can finish the rest of the sheet now :biggrin:

( no doubt i'll be back on in 10 minutes asking about some more differentiating thing :redface:

Reply 7

OP

sooooooooo... y = x^2 - x^-2
dy/dx = 2x + 2x^-3 ???

yeah, that's right

OP

yey

OP

right... onto diff type of qus now... hehe still differentiation tho !
cld u check if im doing this right pleeeeease !?!

find the gradient of the curve with equation y=f(x) at the point where x=a when:
f(x) = (x-4)/(x^2) , a = -2

(i assume that means work out the dx/dy thing then put -2 in instead of x )

ive worked out the dy/dx = -x^-2 + 8x^-3
and so the gradient is -1.25
is any of that right ?

thank you soooo much for all your help.....i havent been taught how to do any of this... im gonna buy a text book this weekend.... you are all wonderful !

Reply 11

m:)ckel

well, dy/dx = -x^-2 + 8x^-3
but i get the same answer as you for the gradient, so i'm assuming your original post was just a typo (you wrote dy/dx = -x^2......)

Reply 12

OP

yeah.. sorry typo, i've changed it now
how do you do y = square root of (12x) , a= square root of 3

argh ! - dont worry, i'll get the hang of this soon - honest ! :eek:

Reply 13

do it the same way you did the others, i.e. taking the exponent out as a constant, and the new exponent is (old exponent - 1)

y = (12^1/2) x^1/2

dy/dx = (12^1/2) (1/2) x^-1/2

at x = 3^1/2

dy/dx = (12^1/2) (1/2) (3^1/2)^-1/2

= root(3) / root(root(3)) = root(root(3))

Reply 14

OP

but how can u do the square root of 12 separate from square root of x and then just times them together...cos isnt the question asking the square root of 12 x ...... or am i TOTALLY missing the point ?and what is root (root3) ?

Reply 15

franks

but how can u do the square root of 12 separate from square root of x and then just times them together...cos isnt the question asking the square root of 12 x ...... or am i TOTALLY missing the point ?

(ab)^n ==== a^n * b^n

Reply 16

OP

right ok...

y = (12^1/2) x^1/2

dy/dx = (12^1/2) (1/2) x^-1/2

what i dont understand is that you seem to have got rid of the first half of the equation ... i think that from y=12^1/2 x^1/2
that the 12^1/2 goes to (1/2)(12^-1/2) in the dy/dx thing
and that the x^1/2 goes to (1/2)(x^-1/2)

it looks like ...... im really confused - can ANYONE explain this???? the question again was y=square root of 12 , what is gradient when x is square root of 3

Reply 17

franks

right ok...

shouldnt it be dy/dx = (1/2)(12^-1/2) (1/2)(x^-1)

no, d/dx{x^1/2} = 1/2 x^(1/2 - 1) = 1/2 x^-1/2

Reply 18

OP

right ok...

y = (12^1/2) x^1/2

dy/dx = (12^1/2) (1/2) x^-1/2

what i dont understand is that you seem to have got rid of the first half of the equation ... i think that from y=12^1/2 x^1/2
that the 12^1/2 goes to (1/2)(12^-1/2) in the dy/dx thing
and that the x^1/2 goes to (1/2)(x^-1/2)

it looks like ...... im really confused - can ANYONE explain this???? the question again was y=square root of 12 , what is gradient when x is square root of 3

Reply 19