Hi! could you please determine the derivative of this function? y=∛(x^4-x-1)/√(5&x^5+3x) Thanks

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The derivative of `y = root(3)(x^4 - x - 1)/sqrt(5x^5 + 3x)` has to be determined.

Using the quotient rule:

y' = `((root(3)(x^4 - x - 1))'*(sqrt(5x^5 + 3x)) - root(3)(x^4 - x - 1)*(sqrt(5x^5 + 3x))')/(sqrt(5x^5 + 3x))^2`

` `= `(2*(4x^3 - 1)*(5x^5 + 3x) - (x^4 - x - 1)*3*(25x^4 + 3))/(6*(5x^5 + 3x)^(3/2)*(x^4 - x - 1)^(2/3))`

= `(2*(20x^8-5x^5+12x^4-3x)-3*(25x^8-25x^5-25x^4+3x^4-3x-3))/(6*(5x^5+3x)^(3/2)*(x^4-x-1)^(2/3))`

= `(40x^8-10x^5+24x^4-6x - 75x^8+75x^5+75x^4-9x^4+9x+9)/(6*(5x^5+3x)^(3/2)*(x^4-x-1)^(2/3))`

= `(-35x^8 +65x^5+90x^4+3x+9)/(6*(5x^5+3x)^(3/2)*(x^4-x-1)^(2/3))`

The derivative of `y = root(3)(x^4 - x - 1)/sqrt(5x^5 + 3x)` is `y'=(-35x^8 +65x^5+90x^4+3x+9)/(6*(5x^5+3x)^(3/2)*(x^4-x-1)^(2/3))`

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