Trail HSc questions (1 Viewer)

[collide]

f(x) = log3(1/x^2)
= log31-log3x^2
= 0 - log3x^2
= -lnx^2/ln3

using quotient rule..

f'(x)=-(2x/x^2).ln3/(ln3)^2
=(-2x.ln3/x^2) x (1/(ln3)^2)
= -2x/x^2.ln3

i think.. lol.

 

[FOB all the way]

sorry i dont follow after the quotient rule

 

[collide]

using quotient rule..
f'(x)= (u'v-v'u)/v^2

let u = -lnx^2
u' = -2x/x^2

let v=ln3
v'=0

therefore

f'(x)=-(2x/x^2).ln3/(ln3)^2
=(-2x.ln3/x^2) x (1/(ln3)^2)
= -2x/(x^2).ln3


sorry that's the best i can explain it.. it might be best if you write it out. hope that helps.

 

[FOB all the way]

sorry why did yo times by 1/in3^2

 

[collide]

sorry why did yo times by 1/in3^2

because i was simplifying it.. multiplying the reciprocal..

for example (1/5) divide by 7
= (1/5) x (1/7)
= 1/35

but in this case

f'(x)=((-2x/x^2).ln3) divide by (ln3)^2
=(-2x.ln3/x^2) x (1/(ln3)^2)
= -2x/(x^2).ln3

is that better? :S

 

[FOB all the way]

thanks heeps

 

[collide]

no worries
is the answer correct though?

 

[FOB all the way]

wont find out till i go back to school