Question

derivada de 2sen(3x)x+3cos(x^2)-x^3

Answer 1

Utilizamos a regra da cadeia para derivadas de funções compostas

$\boxed{\boxed{\dfrac{d}{dx}~f(g(x))=f'(g(x))\cdot g'(x)}}$
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$m(x)=2xsen~(3x)+3cos(x^{2})-x^{3}$

Vamos fazer por partes

Derivando 2x*sen(3x):

Chamemos sen(x) de g(x) e 3x de h(x), ficando com sen(3x) = g(h(x))

$g(h(x))=sen~3x\\\\\\\dfrac{d}{dx}g(h(x))=\dfrac{d}{dx}sen~3x\\\\\\\dfrac{d}{dx}g(h(x))=cos~3x\cdot 1\cdot3x^{1-1}\\\\\\\dfrac{d}{dx}g(h(x))=3cos~3x$

Derivando agora 2x*sen(3x) pela regra do produto:

$\dfrac{d}{dx}2x\cdot sen(3x)=\left(\dfrac{d}{dx}2x\right)\cdot sen~3x+\left(\dfrac{d}{dx}sen(3x)\right)\cdot2x\\\\\\\dfrac{d}{dx}2x\cdot sen(3x)=2\cdot sen(3x)+3cos(3x)\cdot 2x\\\\\\\boxed{\dfrac{d}{dx}2x\cdot sen(3x)=2sen(3x)+6xcos(3x)}$

Derivando 3cos(x²):

$f(x)=3cos(x^{2})\\f'(x)=3[-sen(x^{2})]\cdot2\cdotx^{2-1}\\f'(x)=-3sen(x^{2})\cdot2x\\f'(x)=-6xsen(x^{2})$

Voltando pra m(x):

$m(x)=2xsen(3x)+3cos(x^{2})-x^{3}\\m'(x)=[2sen(3x)+6xcos(3x)]+[-6xsen(x^{2})]-3\cdot x^{3-1}\\m'(x)=2sen(3x)+6xcos(3x)-6xsen(x^{2})-3x^{2}$