Question

How do you differentiate `y=(2^x-x^2)/(1-log_3x)`?

Answer 1

`dy/dx=((1-log_3x)(2^xln2-2x)+(2^x-x^2)/(xln3))/(1-log_3x)^2`

Explanation:

`color(orange)"Reminders"`

`• d/dx(a^x)=a^xlna`

`• d/dx(log_ax)=1/(xlna)`

`"differentiate using the "color(blue)"quotient rule"`

`"Given " y=(g(x))/(h(x))" then"`

`dy/dx=(h(x)g'(x)-g(x)h'(x))/(h(x))^2larr" quotient rule"`

`"g(x)=2^x-x^2rArrg'(x)=2^xln2-2x`

`h(x)=1-log_3xrArrh'(x)=-1/(xln3)`

`rArrdy/dx=((1-log_3x)(2^xln2-2x)-(2^x-x^2)(-1/(xln3)))/(1-log_3x)^2`