Question

se y= 2u3 e u= 5x2+2x-1 determine dy/dx

Answer 1

Regra da cadeia:

$\boxed{\boxed{\dfrac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)}}$
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$y=2u^{3}$

Como u = 5x² + 2x - 1:

$y=2(5x^{2}+2x-1)^{3}$

Derivando y em relação a x, pela regra da cadeia:

$\dfrac{dy}{dx}=2\cdot3(5x^{2}+2x-1)^{3-1}\cdot\dfrac{d}{dx}(5x^{2}+2x-1)\\\\\\\dfrac{dy}{dx}=6(5x^{2}+2x-1)^{2}\cdot(5\cdot2x^{2-1}+2\cdot1x^{1-1}-0)\\\\\\\dfrac{dy}{dx}=6(5x^{2}+2x-1)^{2}\cdot(10x+2)\\\\\\\dfrac{dy}{dx}=6\cdot(5x^{2}+2x-1)^{2}\cdot2\cdot(5x+1)\\\\\\\boxed{\boxed{\dfrac{dy}{dx}=12\cdot(5x^{2}+2x-1)^{2}\cdot(5x+1)}}$
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Podíamos também usar a regra da cadeia com a notação de Leibniz:

$\boxed{\boxed{\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}}}$

Derivando y em relação a u:

$y=2u^{3}\\\\\\\dfrac{dy}{du}=3\cdot2u^{3-1}\\\\\\\boxed{\dfrac{dy}{du}=6u^{2}}$

Derivando u em relação a x:

$u=5x^{2}+2x-1\\\\\\\dfrac{du}{dx}=2\cdot5x^{2-1}+1\cdot2x^{1-1}-0\\\\\\\boxed{\dfrac{du}{dx}=10x+2=2\cdot(5x+1)}$

Então:

$\dfrac{dy}{dx}=6u^{2}\cdot2\cdot(5x+1)=12u^{2}(5x+1)$

Como u = 5x² + 2x - 1:

$\boxed{\boxed{\dfrac{dy}{dx}=12\cdot(5x^{2}+2x-1)^{2}\cdot(5x+1)}}$