Question

Derivada das funções abaixo:

Answer 1

d) 
$f(x) = \frac{g(x)}{h(x)} \\ \\ \boxed{f'(x) = \frac{h(x)\cdot g'(x) - h'(x)\cdot g(x)}{[h(x)]^2}}$

Separadamente:
g(x) = 1 → g'(x) = 0
h(x) = x + 1 → h'(x) = 1

Substituindo:

$f'(x) = \frac{(x+1)\cdot 0 - 1\cdot 1}{(x+1)^2} \\ \\ f'(x) = \frac{0 - 1}{x^2 + 2x + 1} \\ \\ \boxed{f'(x) = - \frac{1}{x^2 + 2x + 1}}$ 


e)
$f(x) = \frac{1}{x^2 + 1} + x^5cosx$

$f(x) = \frac{g(x)}{h(x)} + i(x)\cdot m(x)$

g(x) = 1 → g'(x) = 0
h(x) = x² + 1 → h'(x) = 2x
i(x) = x⁵ → i'(x) = 5x⁴
m(x) = cosx → m'(x) = -sen x


$f'(x) = \frac{h(x)\cdot g'(x) - h'(x)\cdot g(x)}{[h(x)]^2} + i'(x)+ m'(x) \\ \\ f'(x) = \frac{(x^2 + 1)\cdot 0 - 2x\cdot 1}{(x^2 + 1)^2} + 5x^4\ + (-senx) \\ \\ \boxed{f'(x) = - \frac{2x}{(x^2 + 1)^2} - 5x^4- senx}}$

f)

$f'(x) = \frac{h(x)\cdot g'(x) - h'(x)\cdot g(x)}{[h(x)]^2} \\ \\ \boxed{f'(x) = \frac{1}{(x-1)^2} }$

g)

$g'(x) = 0 \\ h'(x) = -senx \\ \\ f'(x) = \frac{0 - (-senx)\cdot 1}{(2 + senx)^2} \\ \\ \boxed{f'(x) = \frac{senx}{(2 + senx)^2} }}$

h)
$g'(x) = 2x + 3 \\ h'(x) = 4x^3 + 2x \\ \\ f'(x) = \frac{h(x)\cdot g'(x) - h'(x)\cdot g(x)}{[h(x)]^2} \\ \\ f'(x) = \frac{(x^4 + x^2 + 1)\cdot(2x + 3) - (4x^3 + 2x)\cdot (x^2 + 3x + 2)}{(x^4 + x^2 + 1)^2} \\ \\ f'(x) = \frac{(2x^5 + 3x^4 + 2x^3 + 3x^2 + 2x + 3) - (4x^5 + 12x^4 + 10x^3 + 6x^2 + 4x)}{(x^4 + x^2 + 1)^2} \\ \\\boxed{ f'(x) = \frac{-2x^5 - 8x^4 - 8x^3 - 3x^2 - 2x + 3}{(x^4 + x^2 + 1)^2} }}}$

i)
$g'(x) = - cosx \\ h'(x) = senx \\ \\ f'(x) = \frac{h(x)\cdot g'(x) - h'(x)\cdot g(x)}{[h(x)]^2} \\ \\ f'(x) = \frac{(2 - cosx)(-cosx) - senx(2 - senx)}{(2- cosx)^2} \\ \\ f'(x) = \frac{(-2cosx + cos^2x) - (2senx + sen^2x)}{(2-cosx)^2}$

Na parte de cima da fração, teremos:

$(-2cosx + cos^2x) - (2senx + sen^2x) \\ \\ -2cox + cos^2x - 2senx + sen^2x \\ \\ \boxed{cos^2x + sen^2x = 1} \\ \\ \boxed{\boxed{-2cos -2sen + 1}}$

Voltando a f'(x):

$\boxed{f'(x) = \frac{-2cos -2sen + 1}{(2-cosx)^2}}}$