Let `f(x)=(6x^3-5)/x^4` , find f'(x)?

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To take the derivative of f(x), apply quotient rule which is `(u/v)' = (v*u' - u*v')/v^2` .

So let,

`u =6x^3-5`              and          `v=x^4`

Then,determine u' and v' .

`u'=18x^2`               and          `v'=4x^3`

And, plug-in u, v, u' and v' to the  formula of quotient rule.

`f'(x) = (x^4*18x^2 - (6x^3-5)*4x^3)/(x^4)^2=(18x^6 - (24x^6-20x^3))/x^8`

`f'(x)= (18x^6-24x^6+20x^3)/x^8=(-6x^6+20x^3)/x^8`

To simplify further, factor out the GCF in the numerator.

`f'(x) = (2x^3(-3x^3+10))/x^8=(2x^3(10-3x^3))/x^8`

And cancel common factor.

`f'(x)= (2(10-3x^3))/x^5`

Hence, the derivative of f(x) is `f'(x)= (2(10-3x^3))/x^5` .

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