The Student Room Group
Reply 1
rewrite it as (6/5)(x^-3) + 4(x^3)
Reply 2
El Stevo
rewrite it as (6/5)(x^-3) + 4(x^3)

what i said then ok cool
Mr. Brightside
basically diff indices
6
----
5x^3 + 4x^3
now the 4x^3 bit is simply 12x squared but the six over 5x^3 is confusing me a bit, would it be just 6/5 times x^3 and then differentiate it. that seems too easy. any suggestions?

d/dx {6/(5x^3) + 4x^3} = d/dx {(6/5)(x^-3) + 4x^3} = (6/5)(-3)(x^-4) + (4)(3)(x^2) = 12x^2 - 18/(5x^4) = (60x^6 - 18)/(5x^4)

Nima
Reply 4
Mr. Brightside
what i said then ok cool


no... you said (6/5)*(x^3)... what i said has a very important difference... (6/5)*(x^-3)
El Stevo
no... you said (6/5)*(x^3)... what i said has a very important difference... (6/5)*(x^-3)

his

6
----
5x^3


is 6/(5x^3) ?
Reply 6
yes... but his
Mr. Brightside
would it be just 6/5 times x^3 and then differentiate it
is wrong...
oh yeh ok then.

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