Question

ME AJUDA GENTEEEE

Calcule a derivada:

a) /=(2^2−3+1)/8^3+2

b) /=cos /1−

c) ′() = ^3√3^2 + 2 + 2

d) /=1/(−5^2+3+4)^4

e) ′() = (2^2 − 10 + 6)(1 − ^−3)

Answer 1

Resposta:

Explicação passo-a-passo:

a)

$\frac{dy}{dx}=\frac{(2x^2-\:3x\:+1)}{8x^3} = \\\\=\frac{\frac{d}{dx}\left(2x^2-3x+1\right)\left(8x^3+2\right)-\frac{d}{dx}\left(8x^3+2\right)\left(2x^2-3x+1\right)}{\left(8x^3+2\right)^2} = \\\\=\frac{\left(4x-3\right)\left(8x^3+2\right)-24x^2\left(2x^2-3x+1\right)}{\left(8x^3+2\right)^2} =\\\\=\frac{32x^4-24x^3+8x -6-48x^4+72x^3-24x^2}{\left(8x^3+2\right)^2} =\\\\=\frac{-16x^4+48x^3-24x^2 +8x -6}{\left(8x^3+2\right)^2}$

b)

$\frac{dy}{dx} = \frac{cosx}{1 - sen x} =\\\\\\ =\frac{\frac{d}{dx}\left(\cos \left(x\right)\right)\left(1-\sin \left(x\right)\right)-\frac{d}{dx}\left(1-\sin \left(x\right)\right)\cos \left(x\right)}{\left(1-\sin \left(x\right)\right)^2}=\\\\=\frac{\left(-\sin \left(x\right)\right)\left(1-\sin \left(x\right)\right)-\left(-\cos \left(x\right)\right)\cos \left(x\right)}{\left(1-\sin \left(x\right)\right)^2}=\\\\= \frac{-sin(x) + sin^2(x) +cos^2(x)}{(1-sin^2(x))^2} =$

$= \frac{1 - sin^2 (x)}{(1 - sin^2(x))^2 } = \\\\= \frac{1 }{(1 - sin(x)) }$

c)

$f'(x) = \sqrt[3]{3x^2+ 2x + 2} \\=\frac{1}{3\left(3x^2+2x+2\right)^{\frac{2}{3}}}\frac{d}{dx}\left(3x^2+2x+2\right)=\\\\=\frac{1}{3\left(3x^2+2x+2\right)^{\frac{2}{3}}}\left(6x+2\right)=\\\\=\frac{6x+2}{3\left\sqrt[3]{(3x^2+2x+2)^2\right)} }}$

d)

$\frac{dy}{dx}\left(\frac{1}{\left(-5x^2+3x\:+\:4\right)^4}\:\:\right)\\\\\\\\\frac{dy}{dx}\left\left(-5x^2+3x+4\right)^{-4}\right\\\\\\=-\frac{4}{\left(-5x^2+3x+4\right)^5}\frac{d}{dx}\left(-5x^2+3x+4\right)=\\\\=-\frac{4}{\left(-5x^2+3x+4\right)^5}\left(-10x+3\right)=\\\\=-\frac{4\left(-10x+3\right)}{\left(-5x^2+3x+4\right)^5}\\\\=-\frac{\left(-40x+12\right)}{\left(-5x^2+3x+4\right)^5}$

e)

$f'(x) = (2x^2 - 10x + 6)(1-t^{-3})=\\ =\left(1-t^{-3}\right)\frac{d}{dx}\left(2x^2-10x+6\right)=\\\\=\left(1-t^{-3}\right)\left(\frac{d}{dx}\left(2x^2\right)-\frac{d}{dx}\left(10x\right)+\frac{d}{dx}\left(6\right)\right)=\\\\=\left(1-t^{-3}\right)\left(4x-10+0\right)=\\=\left(1-t^{-3}\right)\left(4x-10\right)$