I tried differentiating (2x-6) / (x^2 -6x +10) ...and this is what i got:
2*-1(x^2 - 6x +10)^(-1-1) * (2x-6) = -2(2x-6) / (x^2 - 6x +10 )^2
= (-4x + 12) / (x^2 -6x +10)^2
why isn't this correct?? Show the correct working...
your usage of chain rule is wrong (although some would prefer to use quotient rule here)
d/dx (2x-6) / (x^2 -6x +10)
= d/dx (2x-6) (x^2 -6x +10)^(-1)
= (2x-6)(-1)(x^2 -6x +10)^(-2)(2x-6) + (x^2 -6x +10)^(-1)(2)
Try continuing from the above.
Regards,
Eagle
(ExamWorld)
Hi,
It is interesting to find the integral of (2x - 6)/(x^2 - 6x + 10) with respect to x, which is
ln |x^2 - 6x + 10| + c. Thanks :)
Cheers,
Wen Shih
Ok yeah, the chain rule is not the right thing to apply here!!
@ eagle. I am not sure but your usage of the quoient rule seems incorrect.
let y = (2x-6) / (x^2 -6x +10)
from the quoient rule equation in the text book:
if y = u /v then dy / dx = ( (du/dx) v - u (dv / dx) ) / (v)^2
therefore in my case u=2x - 6 and v= (x^2 - 6x +10). So wouldn't it be:
dy/dx = ( 2(x^2 - 6x +10) - (2x -6)(2x -6) ) / (x^2 - 6x +10)^2
= ( 2x^2 - 12x +20 - (4x^2 -24x +36) ) / (x^2 - 6x +10)^2
= ( 2x^2 - 12x +20 - 4x^2 +24x -36) / (x^2 - 6x +10)^2
= ( -2x^2 + 12x -16) / (x^2 - 6x +10)^2
Is this correct?? We don't have to worry about (x^2 -6x +10)^(-1) in this situation, right???
actually I meant product rule... together with chain rule... Sorry for not stating that term...
Product rule and Quotient rule can be used interchangeably; it's just how you write the terms. My method was product rule :D