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

Question

Aurora Alvarado · Accepted Answer

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

Isabella Ackerman · Answer

undefined To differentiate y=(2^x-x^2)/(1-log_3x), apply the quotient rule. The derivative is given by the formula:

y' = [(2^x * ln(2) - 2x) * (1 - log_3(x)) - (2^x - x^2) * (-1/(x * ln(3))))] / (1 - log_3(x))^2

Axel Alvarez · Answer

undefined To differentiate $ y = \frac{2^x - x^2}{1 - \log_3{x}} $, you can use the quotient rule and the chain rule where necessary. The quotient rule states that for functions $ u(x) $ and $ v(x) $, if $ y = \frac{u(x)}{v(x)} $, then $ y' = \frac{u'v - uv'}{v^2} $.

First, find the derivatives of the numerator $ u(x) = 2^x - x^2 $ and the denominator $ v(x) = 1 - \log_3{x} $:

$$ u'(x) = \frac{d}{dx}(2^x - x^2) = 2^x \ln{2} - 2x $$
$$ v'(x) = \frac{d}{dx}(1 - \log_3{x}) = -\frac{1}{x \ln{3}} $$

Now, apply the quotient rule:

$$ y' = \frac{(2^x \ln{2} - 2x)(1 - \log_3{x}) - (2^x - x^2)(-\frac{1}{x \ln{3}})}{(1 - \log_3{x})^2} $$

This expression represents the derivative of $ y $ with respect to $ x $.