Cindelyn

Como você diferencia # y = log x ^ 2 #? Responda: #dy/dx=2/x# Explicação: There are 2 possible approaches. #color(blue)”Approach 1″# differentiate using the #color(blue)”chain rule”# #color(red)(bar(ul(|color(white)(2/2)color(black)(d/dx(log(f(x)))=1/(f(x)).f'(x))color(white)(2/2)|)))# #y=log(x^2)# #rArrdy/dx=1/x^2.d/dx(x^2)=1/x^2 xx2x=2/x# #color(blue)”Approach 2″# Using the #color(blue)”law of logs”# then differentiate. #color(orange)”Reminder ” color(red)(bar(ul(|color(white)(2/2)color(black)(logx^n=nlogx)color(white)(2/2)|)))# #y=logx^2=2logx# #rArrdy/dx=2xx 1/x=2/x#