How to solve?? (1 Viewer)

[epicFAILx]

Im starting differentiation in maths... quotient rule. ANd a question i have come across is:

f(x)= 1/ x^2+2x-3


It looks okay I guess. But what would be the smarter way of solving it.
Factorising then doing the quotient rule
Or just doing the quotient rule and simplifying?

 

[RivalryofTroll]

Im starting differentiation in maths... quotient rule. ANd a question i have come across is:

f(x)= 1/ x^2+2x-3


It looks okay I guess. But what would be the smarter way of solving it.
Factorising then doing the quotient rule
Or just doing the quotient rule and simplifying?

I think you should just use dy/dx = vu' - uv'/v^2 like normal =/
I'm not 100% sure (I'm only in Year 10 xD so yeah)

EDIT:
x^2-5/(x^2+2x-3)^2 ?
What did you get? What's the textbook answer?

 

[brettymaccc]

I don't think there's a 'smarter way' of solving it as especially early on, they throw ugly things at you to differentiate as you can be asked to differentiate anything later on.

Just for good practice, basically.

EDIT: Although, it could be sped up if you were to realise that u' = 0, and thus vu' = 0.

Thus, the formula becomes -uv'/v^2, or just -v'/v^2 (as u = 1.)

So it becomes -(2x + 2)/(x^2+2x-3)^2, or -2(x+1)/(x^2+2x-3)^2.

 

[fiesycal]

Is that 1/(x^2+2x-3) or something else? If it's the former you don't need to use quotient rule as there is no x on the top. you can just change the form to (x^2+2x-3)^-1 and differentiate it from there.

 

[SpiralFlex]

It would be better to change it to index form and use the chain rule.

(Watch for constants in numerators!)

 

[epicFAILx]

It would be better to change it to index form and use the chain rule.

(Watch for constants in numerators!)

Ahhh it makes sense now. I just made a stupid mistake with the 1/ at the beginning. Thanks Spriral.

Umm. One more thing. Tangents and Normals. Im not to sure about.

Like eg: a simple y=x^2 (finding the gradient of the tangent at x=2)
do you derive it first then substitute for x

or something like that?

 

[epicFAILx]

I think you should just use dy/dx = vu' - uv'/v^2 like normal =/
I'm not 100% sure (I'm only in Year 10 xD so yeah)

EDIT:
x^2-5/(x^2+2x-3)^2 ?
What did you get? What's the textbook answer?

I dont know the answer.. its not from a textbook

My answer origianlly was x+2x-3 x 2x-2/ (x^2+ 2x -3)


Lol.

 

[Alkanes]

Spi beat me to it again....

 

[RivalryofTroll]

I dont know the answer.. its not from a textbook

My answer origianlly was x+2x-3 x 2x-2/ (x^2+ 2x -3)


Lol.

I just realized I totally did it wrong xD
Yeah its (x^2+2x-3).0 (thought it was .1) - 1.(2x+2)/x^2+2x-3/(x^2+2x-3)^2
= -2x-2/(x^2+2x-3)^2

 

[RivalryofTroll]

Ahhh it makes sense now. I just made a stupid mistake with the 1/ at the beginning. Thanks Spriral.

Umm. One more thing. Tangents and Normals. Im not to sure about.

Like eg: a simple y=x^2 (finding the gradient of the tangent at x=2)
do you derive it first then substitute for x

or something like that?

dy/dx = 2x
= 4 ? Oo" You mean this?