Question

Determine o valor das seguintes derivadas:

$l) y = sin (4x^{2} +2)\\\\m) f(x) = x sin x - x^{2} \\\\n) f(x) = \sqrt{cos x - (tan x)^{3} } \\\\o) f(x) = (x^{2} + 4x)^{\frac{5}{4} } \\\\p) f(x) = 2\sqrt{x(x^{2} +2x)^{\frac{3}{4} } }$

Answer 1

Resposta:

Explicação passo-a-passo:

l)

$y=sin(4x^2+2) \\y' =\cos \left(4x^2+2\right)\frac{d}{dx}\left(4x^2+2\right)\\y' =8x . \cos \left(4x^2+2\right)$

m)

$f(x) = xsinx - x^2\\f'(x) =\frac{d}{dx}\left(x\sin \left(x\right)\right)-\frac{d}{dx}\left(x^2\right)\\f'(x) =\sin \left(x\right)+x\cos \left(x\right)-2x$

n)

$f(x) = \sqrt{cos(x) -(tan (x))^3} \\f'(x) =\frac{1}{2\sqrt{\cos \left(x\right)-\left(\tan \left(x\right)\right)^3}}\frac{d}{dx}\left(\cos \left(x\right)-\left(\tan \left(x\right)\right)^3\right)\\f'(x) =\frac{1}{2\sqrt{\cos \left(x\right)-\left(\tan \left(x\right)\right)^3}}\left(-\sin \left(x\right)-3\tan ^2\left(x\right)\sec ^2\left(x\right)\right)\\f'(x) =\frac{-\sin \left(x\right)-3\tan ^2\left(x\right)\sec ^2\left(x\right)}{2\sqrt{\cos \left(x\right)-\tan ^3\left(x\right)}}$

o)

$\frac{d}{dx}\left(\left(x^2+4x\right)^{\frac{5}{4}}\right)=\\=\frac{5\left(x^2+4x\right)^{\frac{1}{4}}}{4}\left(2x+4\right) =\\=\frac{5\left(x^2+4x\right)^{\frac{1}{4}}}{4}\left(2x+4\right)=\\ =\frac{5\left(x+2\right)\left(x^2+4x\right)^{\frac{1}{4}}}{2}$

p)

$f(x) = 2\sqrt{x(x^2+2x)^{\frac{3}{4} } } \\f'(x) =2\frac{d}{dx}\left(\sqrt{x\left(x^2+2x\right)^{\frac{3}{4}}}\right)\\\\f'(x) =2\cdot \frac{1}{2\left(x\left(x^2+2x\right)^{\frac{3}{4}}\right)^{\frac{1}{2}}}\cdot \frac{5x^2+7x}{2\left(x^2+2x\right)^{\frac{1}{4}}}\\f'(x) =\frac{x^{\frac{1}{2}}\left(5x+7\right)}{2\left(x^2+2x\right)^{\frac{5}{8}}}$