Question

Qual a derivada da função:
g(x) = (x^2+3) . ln(1+cos2x)

Answer 1

Será preciso utilizar as Regras do Produto e da Cadeia:

$\boxed{\dfrac{\mathrm{d}}{\mathrm{d}x}\left[f(x)g(x)\right]=g(x)\dfrac{\mathrm{d}}{\mathrm{d}x}\left(f(x)\right)+f(x)\dfrac{\mathrm{d}}{\mathrm{d}x}\left(g(x)\right)}$

$\boxed{\dfrac{\mathrm{d}}{\mathrm{d}x}\left[f(g(\cdots x))\right]=\dfrac{\mathrm{d}}{\mathrm{d}x}\left[f(g(\cdots x))\right]+\dfrac{\mathrm{d}}{\mathrm{d}x}\left[g(\cdots x)\right]+\cdots}$

Temos que:

$\boxed{g(x)=\left(x^2+3\right)\ln{\left(1+\cos{2x}\right)}}$

Sua derivada será dada por:

$\dfrac{\mathrm{d}g}{\mathrm{d}x}=\ln{\left(1+\cos{2x}\right)}\dfrac{\mathrm{d}}{\mathrm{d}x}\left(x^2+3\right)+(x^2+3)\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\ln{\left(1+\cos{2x}\right)}\right)$

$\dfrac{\mathrm{d}g}{\mathrm{d}x}=2x\ln{\left(1+\cos{2x}\right)}+\left(x^2+3\right)\left(\dfrac{1}{1+\cos{2x}}\right)\dfrac{\mathrm{d}}{\mathrm{d}x}\left(1+\cos{2x}\right)\dfrac{\mathrm{d}}{\mathrm{d}x}\left(2x\right)$

$\dfrac{\mathrm{d}g}{\mathrm{d}x}=2x\ln{\left(1+\cos{2x}\right)}+\dfrac{\left(x^2+3\right)(0-\sin{2x})(2)}{1+\cos{2x}}$

$\dfrac{\mathrm{d}g}{\mathrm{d}x}=2x\ln{\left(1+2\cos^2{x}-1\right)}-2\left(x^2+3\right)\left(\dfrac{2\sin{x}\cos{x}}{1+2\cos^2{x}-1}\right)$

$\boxed{\dfrac{\mathrm{d}g}{\mathrm{d}x}=2x\ln{\left(2\cos^2{x}\right)}-2\left(x^2+3\right)\tan{x}}$