Question

f(x)= 2sen2(x)+2cos2(x)+x

Answer 1

Resposta:

f'(x)=4ens+4cos+1

Resposta

$f(x) = 2sen2(x) + 2cos2(x) + x = \\ f(x) = 2sen \times 2x + 2cos \times 2x + x \\ f(x) \frac{d}{dx}(2sen \times 2x + 2cos \times 2x + x) \\ 2sen \times 2x \\ 4senx \\ f(x) = \frac{d}{dx}(2sen \times 2x + 2cos \times 2x + x) \\ f(x) = \frac{d}{dx}(4senx + 2cos \times 2x + x) \\f(x) = \frac{d}{dx}(2sen \times 2x + 2cos \times 2x + x) \\ 2cos \times 2x \\ 4cosx \\ f(x) = \frac{d}{dx}(2sen \times 2x + 2cos \times 2x + x) \\ f(x) = \frac{d}{dx}(4senx + 4cosx + x) \\ f(x) = \frac{d}{dx}(4senx) + \frac{d}{dx}(4cosx) + \frac{d}{dx}(x) \\ \frac{d}{dx}(4senx) \\ 4sen \\ f(x) = \frac{d}{dx}(4senx) + \frac{d}{dx}(4cosx) + \frac{d}{dx} (x) \\ f(x) = 4sen + \frac{d}{dx}(4cosx) + \frac{d}{dx}(x) \\ f(x) = \frac{d}{dx}(4senx) + \frac{d}{dx}(4cosx) + \frac{d}{dx}(x) \\ \frac{d}{dx}(4cosx) \\ 4cos \\ f(x) = \frac{d}{dx}(4senx) + \frac{d}{dx}(4cosx) + \frac{d}{dx}(x) \\ f(x) = 4sen + 4cos + \frac{d}{dx}(x) \\ f(x) = \frac{d}{dx}(4senx) + \frac{d}{dx}(4cosx) + \frac{d}{dx}(x) \\ \frac{d}{dx}(x) \\1\\ f(x) = \frac{d}{dx}(4senx) + \frac{d}{dx}(4cosx) + \frac{d}{dx}(x) \\f(x) = 4sen + 4cos + 1 \\ f(x) = 4ens + 4cos + 1 \\ resposta \\ f(x) = 4ens + 4cos + 1$