Como calcular a derivada de y=ln(x-1)cosx/x²+1

Devemos encontrar a derivada da função dada e para isso utilizaremos a propriedade de Regra do Quociente de derivadas:

\(\begin{array}{l} \frac{d}{{dx}}\ln (x - 1)\frac{{\cos x}}{{{x^2} + 1}} = \frac{{({x^2} + 1)\left( {\frac{d}{{dx}}\ln (x - 1)\cos x} \right) - \ln (x - 1)\cos x\left( {\frac{d}{{dx}}{x^2} + 1} \right)}}{{{{({x^2} + 1)}^2}}}\\ \frac{d}{{dx}}\ln (x - 1)\frac{{\cos x}}{{{x^2} + 1}} = \frac{{({x^2} + 1)\left( {\frac{d}{{dx}}\ln (x - 1)\cos x + \ln (x - 1)\left( {\frac{d}{{dx}}\cos x} \right)} \right) - \ln (x - 1)\cos x\left( {\frac{d}{{dx}}{x^2} + 1} \right)}}{{{{({x^2} + 1)}^2}}}\\ \frac{d}{{dx}}\ln (x - 1)\frac{{\cos x}}{{{x^2} + 1}} = \frac{{({x^2} + 1)\left( {\frac{{\cos x}}{{x - 1}} - \ln (x - 1)senx} \right) - \ln (x - 1)\cos x\left( {\frac{d}{{dx}}{x^2} + 1} \right)}}{{{{({x^2} + 1)}^2}}}\\ \frac{d}{{dx}}\ln (x - 1)\frac{{\cos x}}{{{x^2} + 1}} = \frac{{({x^2} + 1)\left( {\frac{{\cos x}}{{x - 1}} - \ln (x - 1)senx} \right) - 2x\ln (x - 1)\cos x}}{{{{({x^2} + 1)}^2}}} \end{array} \)

Portanto, a derivada da função dada será \(\begin{array}{l} \frac{d}{{dx}}\ln (x - 1)\frac{{\cos x}}{{{x^2} + 1}} = \frac{{({x^2} + 1)\left( {\frac{{\cos x}}{{x - 1}} - \ln (x - 1)senx} \right) - 2x\ln (x - 1)\cos x}}{{{{({x^2} + 1)}^2}}} \end{array} \).