Come si differenzia # y = log x ^ 2 #?
Risposta:
#dy/dx=2/x#
Spiegazione:
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#