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Differentiate this (1 Viewer)

annabackwards

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Use the quotient rule
Let u = (x+5)^4
u' = 4(x+5)^3 x 1 = 4(x+5)^3

v = (x-3)
v' = 1

Now y' = vu' - uv'/v^2
= [ (x-3) x 4(x+5)^3 - (x+5)^4 ] / (x-3)^2
Then just simplify :)
 

Shadowdude

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That'd be a function of a function and a quotient rule:

[(x+5)^4] / (x-3)

uv - vu/v^2

u = (x+5)^4
v = x-3
u' = 4x(x+5)
v' = 1

vu' - uv' / v^2
4x(x-3)(x+5) - (x+5)^4 / (x-3)^2

I believe that simplifies down/complicates up to:

f'x = (3x^4+28x^3-30x^2-900x-2125)/(x^2-6x+9)
 

annabackwards

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That'd be a function of a function and a quotient rule:

[(x+5)^4] / (x-3)

uv - vu/v^2

u = (x+5)^4
v = x-3
u' = 4x(x+5)
v' = 1

vu' - uv' / v^2
4x(x-3)(x+5) - (x+5)^4 / (x-3)^2

I believe that simplifies down/complicates up to:

f'x = (3x^4+28x^3-30x^2-900x-2125)/(x^2-6x+9)
Take care when using the chain rule :)
 

Shadowdude

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I knew I got something wrong... well, looks like you helped two people anna!
 

hscishard

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Im afraid i have another question. Differentiate

y = (e^px + e^-qx) / r


I used the quotient rule, but the answers show:

y' = pe^px - qe^-qx / r

How did they get that answer?
 
Last edited:

fingerman

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y= e^px + e^-qx / r

using quotient rule

y'= (pe^px - qe^-qx)r / r.r

therefore y' = pe^px - qe^-qx / r
 

hscishard

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Huh? This is how i did it.

r(pe^px - qe^-qx) - 1(e^px + e^-qx) / r^2

If I need to expand that to get the answer, i'll get the "Wow" feeling because i tried that and couldn't solve it in any way.

I think how you got just r(pe^px - qe^qx) / r^2 is by saying the differentiate of "r" is 0. But isnt "r" a variable?
 

hscishard

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That'd be a function of a function and a quotient rule:


I believe that simplifies down/complicates up to:

f'x = (3x^4+28x^3-30x^2-900x-2125)/(x^2-6x+9)
The teacher's answer turned out to be (3x-17)(x+5)^3 / (x-3)^2. They factorised and simplified the "-4(x-3)(x+5)^3 - (x+5)^4 / (x-3)^2". God bloody coaching teachers.
 
Last edited:

philphie

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i don't know how you people can stand typing equations, it takes me fucking forever!!!!
 

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