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#

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