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

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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

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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 ).

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