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Differentiation

How would I go about differentiating this:

(4a - 3) over root of a.

I know how to differentiate but I've got absolutely no idea how to turn the above into the form x to the power of something.

Reply 1

fL3X

4a/(roota) - 3/(roota)

y=4 root a - 3^ (a)-1/2

y= 4 a^1/2 - 3 a^-1/2

dy/dx=...try it

Reply 2

stattotheblade

Use the quotient rule. From C3

Is this the equation you are speaking of

f(a) = (4a-3)/(√a) or in other terms (4a-3)/(a^(1/2))

For this you'd use this rule

f'(a) = [(Bottom x diff of top) - (Top x diff of bottom)]/(bottom ^ 2)

So in this case

f'(a) = [(4a - 3) x *(0.5 x a ^ -0.5)* - (√a) x 4] / a

Does that help.

NB - bit with stars is the chain rule

Reply 3

Chaoslord

fL3X

4a/(roota) - 3/(roota)

y=4 root a - 3^ (a)-1/2

y= 4 a^1/2 - 3 a^-1/2

dy/dx=...try it

should be 3a^1/2

latex this bad boy

y = \frac{4a-3}{\sqrt{a}}

y = \frac{4a}{\sqrt{a}} - \frac{3}{\sqrt{a}}

y = 4a^{\frac{1}{2}} - 3a^{\frac{-1}{2}}

Reply 4

fL3X

Chaoslord

should be 3a^1/2

latex this bad boy

y = \frac{4a-3}{\sqrt{a}}

y = \frac{4a}{\sqrt{a}} - \frac{3}{\sqrt{a}}

y = 4a^{\frac{1}{2}} - 3a^{\frac{-1}{2}}

We both got the same thing :confused:

Don't know how to use LateX, mate.

Reply 5

Chaoslord

fL3X

We both got the same thing :confused:

Don't know how to use LateX, mate.

yea i know you just made a simple typo you method is 100% correct, i was just making it easier for the OP to see what you was saying, hence "latex this bad boy" i just rewrote what you had already done (and i fixed the typo)

on the maths forum nearish the top of each thread is a latex tutorial, pretty simple once you get the hang out it