The initial point to do is to create the item guideline formula for our referral:
$$ frac {d} {dx} (uv) = uv’ + vu’$$
We have 2 multipliers in the offered feature f( x). The initial multiplicand is $latex sqrt[5] {x ^ 3} $ as well as the various other is $latex (x ^ 5 + 3x ^ 2– 4x)$.
For that reason, we have
$ latex u = sqrt[5] {x ^ 3} $
$ latex v = (x ^ 5 + 3x ^ 2– 4x)$
$ latex f( x) = uv$
Now, we can make use of the item guideline formula to obtain our offered trouble:
$ latex f'( x) = uv’ + vu’$
$$ frac {d} {dx} f( x) = u cdot frac {d} {dx} (v) + v cdot frac {d} {dx} (u)$$
$$ frac {d} {dx} f( x) = (sqrt[5] {x ^ 3}) cdot frac {d} {dx} (x ^ 5 + 3x ^ 2– 4x)+ (x ^ 5 + 3x ^ 2– 4x) cdot frac {d} {dx} (sqrt[5] {x ^ 3} )$$
For any kind of extreme, it is a good idea to reword them in fractional exponent kind:
$$ frac {d} {dx} f( x) = (x ^ 3) ^ {frac {1} {5}} cdot frac {d} {dx} (x ^ 5 + 3x ^ 2– 4x)+ (x ^ 5 + 3x ^ 2– 4x) cdot frac {d} {dx} (( x ^ 3) ^ {frac {1} {5}} )$$
$$ frac {d} {dx} f( x) = (x ^ 3) ^ {frac {1} {5}} cdot (5x ^ 4 + 6x– 4)+ (x ^ 5 + 3x ^ 2– 4x) cdot (frac {1} {5} cdot (x ^ 3) ^ {-frac {4} {5}} cdot 3x ^ 2)$$
Simplifying algebraically, we obtain
$$ f'( x) = x ^ {frac {3} {5}} cdot (5x ^ 4 + 6x– 4)+ (x ^ 5 + 3x ^ 2– 4x) cdot (frac {3} {5} x ^ {-frac {2} {5}} )$$
$$ f'( x) = 5x ^ {frac {23} {5}} + 6x ^ {frac {8} {5}}– 4x ^ {frac {3} {5}} + frac {3} {5} x ^ {frac {23} {5}} + frac {9} {5} x ^ {frac {8} {5}}– frac {12} {5} x ^ {frac {3} {5}} $$
$$ f'( x) = frac {28} {5} x ^ {frac {23} {5}} + frac {39} {5} x ^ {frac {8} {5}}– frac {32} {5} x ^ {frac {3} {5}} $$
And the last solution is:
$$ f'( x) = frac {28x ^ {frac {23} {5}} + 39x ^ {frac {8} {5}}– 32x ^ {frac {3} {5}}} {5} $$
Or using radicals,
$$ f'( x) = frac {28sqrt[5] {x ^ {23}} + 39sqrt[5] {x ^ 8}– 32sqrt[5] {x ^ 3}} {5} $$