Question

Derive:

Questões sobre derivadas

Answer 1

Resposta:

Explicação passo-a-passo:

a) f(x) = \frac{2x^4}{x - 1} \\f'(x) = 2(\cdot \frac{4x^3\left(x-1\right)-1\cdot \:x^4}{\left(x-1\right)^2}) =\\= \frac{2\cdot(4x^4\left\right-\:3x^3)}{\left(x-1\right)^2}

$b) f(x) = 3^{2x +1} +x^{5} - 3x^{4} + 10\\\\f'(x) =\frac{d}{dx}\left(3^{2x+1}\right)+\frac{d}{dx}\left(x^5\right)-\frac{d}{dx}\left(3x^4\right)+\frac{d}{dx}\left(10\right)=\\\\=2\ln \left(3\right)\cdot \:3^{2x+1}+5x^4-12x^3+0\\\\=5x^4-12x^3+2\ln \left(3\right)\cdot \:3^{2x+1}$

$c) y = ln(3x+2) + e^{2x^{2} } \\\\y' =\frac{d}{dx}\left(\ln \left(3x+2\right)\right)+\frac{d}{dx}\left(e^{2x^2}\right)=\\\\=\frac{3}{3x+2}+4x\cdot e^{2x^2}$

$d) f(x) = (x^7 - 3x) . (5x^2 + 2x + 3)\\\\f'(x) =\frac{d}{dx}\left(x^7-3x\right)\left(5x^2+2x+3\right)+\frac{d}{dx}\left(5x^2+2x+3\right)\left(x^7-3x\right)=\\\\=\left(7x^6-3\right)\left(5x^2+2x+3\right)+\left(10x+2\right)\left(x^7-3x\right)=\\\\= 35x^8 + 14x^7 +21X^6- 15X^2-6X-9 + 10X^8 -30X^2+2X^7-6X=\\\\=45X^8+16X^7+21X^6-45X^2-12X-9$

$e) y = 30x^{-20} -4x^{5} -3x+10\\\\y'=\frac{d}{dx}\left(30x^{-20}\right)-\frac{d}{dx}\left(4x^5\right)-\frac{d}{dx}\left(3x\right)+\frac{d}{dx}\left(10\right)=\\\\=-600x^{-21} -20x^{4} -3$

ESPERO TER AJUDADO!!!