Question

Determine o valor das seguintes derivadas:

$a) y = sin (x^{2} ) \\ b) f(x) = e^{x2} \\ c) f(x) = \sqrt[]{sin (x)} \\ \\ d) f(x) =\frac{2}{(2+inx^{2}) } \\ \\\\ e) f(t) =\frac{1}{(t^{4}+1)^{3}}$

Answer 1

Resposta:

Explicação passo-a-passo:

a)

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

b)

$f(x) =e^{x^{2} } \\f'(x) = e^{x^2}\frac{d}{dx}\left(x^2\right)\\f'(x) = 2x. e^{x^2}$

c)

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

d)

$f(x) = \frac{2}{(2+lnx^2)} \\f'(x) =2\frac{d}{dx}\left(\frac{1}{2+\ln \left(x^2\right)}\right)\\f'(x) = 2\frac{d}{dx}\left(\left(2+\ln \left(x^2\right)\right)^{-1}\right)\\f'(x) = 2\left(-\frac{1}{\left(2+\ln \left(x^2\right)\right)^2}\cdot \frac{2}{x}\right)\\f'(x) =-\frac{4}{x\left(2+2\ln \left(x\right)\right)^2}$

e)

$f(t) = \frac{1}{(t^4-1)^3} \\f(t) = \left(\left(t^4-1\right)^{-3}\right)\\f'(t) = -\frac{3}{\left(t^4-1\right)^4}\cdot \:4t^3\\f'(t) =-\frac{12t^3}{\left(t^4-1\right)^4}$