Question

qual derivada da função f(x) = (3x^5 -1) (2 - x^4)

Answer 1

Existem dois jeitos

$\bullet$ Expandir a expressão de $f(x)$:

$f(x)=(3x^{5}-1)(2-x^{4})\\\\f(x)=3x^{5}\cdot2-3x^{5}\cdot x^{4}-1\cdot2+1\cdot x^{4}\\\\f(x)=6x^{5}-3x^{9} - 2 + x^{4}$

Derivando, temos:

$f'(x)=6\frac{d}{dx}x^{5}-3\frac{d}{dx}x^{9}-0+\frac{d}{dx}x^{4}\\\\f'(x)=6\cdot5x^{4}-3\cdot9x^{8}+4x^{3}\\\\\boxed{\boxed{f'(x)=-27x^{8}+30x^{4}+4x^{3}}}$
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$\bullet$ Usar a regra do produto

$f(x)=(3x^{5}-1)(2-x^{4})\\\\f'(x)=\frac{d}{dx}[(3x^{5}-1)(2-x^{4})]\\\\f'(x)=(2-x^{4})\frac{d}{dx}(3x^{5}-1)+(3x^{5}-1)\frac{d}{dx}(2-x^{4})\\\\f'(x)=(2-x^{4})(3\cdot5x^{4}-0)+(3x^{5}-1)(0-4x^{3})\\\\f'(x)=15x^{4}\cdot(2-x^{4})-4x^{3}\cdot(3x^{5}-1)\\\\f'(x)=30x^{4}-15x^{8}-12x^{8}+4x^{3}\\\\\boxed{\boxed{f'(x)=-27x^{8}+30x^{4}+4x^{3}}}$