Question

Determine o valor das seguintes derivadas:

$f) (x) = (x^{2} +1) (x^{2} +2)^{\frac{1}{3} } \\\\g) f(x) = a^{2} + cos^{3} (x)\\\\h) f(x) = e^{-5x} + cos (3x)\\\\i) f(t) = t^{2} + cos t\\\\\\j) f(t) = \frac{cos (t)}{t} \\\\k) f(x) = In(sin x)$

Answer 1

Resposta:

Explicação passo-a-passo:

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

$f'(x) = \frac{2x(x^2+2)^{\frac{1}{3}}3(x^2+2)^{\frac{2}{3} } }{3(x^2+2)^{\frac{2}{3} } } +\frac{2x(x^2+1)}{3(x^2+2)^{\frac{2}{3} } } \\f'(x) = \frac{6x(x^2+2)+2x(x^2+1)}{3(x^2+2)^{\frac{2}{3} }} \\f'(x) = \frac{6x^3 + 12x+2x^3+ 2x}{3(x^2+2)^{\frac{2}{3} }} \\\\f'(x) = \frac{8x^3 + 14x}{3(x^2+2)^{\frac{2}{3} }}$

$g) f(x) = a^2 + cos^3(x)\\f'(x) =\frac{d}{dx}\left(a^2\right)+\frac{d}{dx}\left(\cos ^3\left(x\right)\right)\\f'(x) =-3\cos ^2\left(x\right)\sin \left(x\right)$

$h) f(x) = e^{-5x} + cos(3x)\\f'(x) = \frac{d}{dx}\left(e^{-5x}\right)+\frac{d}{dx}\left(\cos \left(3x\right)\right)\\f'(x)=-5e^{-5x}-\sin \left(3x\right)\cdot \:3$

$i) f(t) =t^2 + cos(t)\\f'(t)=\frac{d}{dt}\left(t^2\right)+\frac{d}{dt}\left(\cos \left(t\right)\right)\\f'(t)=2t-\sin \left(t\right)$

$j) f(t) = \frac{cos(t)}{t} \\f'(t)=\frac{\frac{d}{dt}\left(\cos \left(t\right)\right)t-\frac{d}{dt}\left(t\right)\cos \left(t\right)}{t^2}\\f'(t)=\frac{\left(-\sin \left(t\right)\right)t-1\cdot \cos \left(t\right)}{t^2}\\f'(t) =\frac{\left(-\sin \left(t\right)\right)t-\cos \left(t\right)}{t^2}$

$k) f(x) = ln(sin(x))\\f'(x) =\frac{1}{\sin \left(x\right)}\frac{d}{dx}\left(\sin \left(x\right)\right)\\f'(x)=\frac{1}{\sin \left(x\right)}\cos \left(x\right)\\f'(x) = \frac{cos(x)}{sin(x)} \\f'(x) = cotg(x)$