differentiation help.. (1 Viewer)

[peterkim95]

ive been working on this for over an hr, and i still coodn't get it..:mad1:
A = x ( 5 + 320/(x-4) )
could you please differentiate this?
thank you

 

[watatank]

A = x ( 5 + 320/(x-4) )

A = x ( 5 + 320/(x-4) )

= 325x/(x-4)

dA/dx = [ 325(x-4) - 325 ]/ (x-4)^2

= [ 325 (x-4-1) ] / (x-4)^2

= [ 325 (x-5) ] / (x-4)^2

 

[aussiechica7]

i don't think ur supposed to add the 320 and 5

this is what i would do:


A = x ( 5 + 320/(x-4) )
=5x + 320x/(x-4)
= 5x + (320(x-4) + 1280)/(x-4)
= 5x + 320 + 1280/(x-4)
=5x + 320 + 1280((x-4)^-1)

dA/dx = 5 – 1280/((x-4)^2)

 

[watatank]

i don't think ur supposed to add the 320 and 5

Perhaps the grouping symbols should have been better placed to not mislead.

I read it again and thus did the question again and I got the same answer. Although I prefer differentiating quotients using formula as follows...

if y = u/v

dy/dx = ( vu' - uv' ) / v^2

 

[aussiechica7]

you forgot the 5x.
Using the d(u/v)/dx = (vu' - uv')/(v^2)...

A = x ( 5 + 320/(x-4) )

= 5x + 320x/(x-4)
= 5 + ((x-4)(320)- (320x)(1))/((x-4)^2)
= 5 + (320x- 1280 -320x)/((x-4)^2)
= 5 -1280/((x-4)^2)

 

[peterkim95]

lol, i attempted the question again, and i got it straight after i postd this thread. thanks so much anyway.
** i got another differentiation question.
its (1/2)k . (25-10k) to the power of 1/2
thx pplz:wave:

 

[ssglain]

i didnt forget the 5 ... i simply added 320 and 5 in the start changing
A = x [ 5 + 320/(x-4) ] to A=x [325/x-4]
whats wrong with that? and i just checked my answer with graphmatica which seems correct
and wats ur logic behind not adding 5 and 320 in the start?

You can't add 5 & 320 like that when they're not on the same denominator. If you insist upon addition then:
x[ 5 + 320/(x-4) ] = x[ (5x + 300)/(x - 4) ] = (5x^2 + 300x)/(x - 4)
to be differented with quotient rule. It doesn't make life any easier.

It's not your fault though. That equation is easily misread in a typeset like that. I would have also made the same error if not for a second look.

Just wondering, can we type equations in maths typesets here in the forum?

[Edit:]

i got another differentiation question.

its (1/2)k . (25-10k) to the power of 1/2

thx pplz

I got the same answer as k02033 had. This is the working out if you needed:
y = (1/2)k*(25-10k)^1/2

y'= [(1/2)*(25-10k)^(1/2)] + [(1/2)k*(1/2)(25-10k)^(-1/2)*(-10)]

= [(1/2)*(25-10k)^(1/2)] - [(5/2)k*(25-10k)^(-1/2)]

= [(1/2)*(25-10k)^(-1/2)] * [(25-10k) - 5k]

= (25-15k)/[2*(25-10k)^1/2]