How Do You Differentiate f(x,y) = e^(2x) * ln(x/y^2) with Respect to x?

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The differentiation of the function f(x,y) = e^(2x) * ln(x/y^2) with respect to x involves applying the product rule and treating y as a constant. The derivative is calculated as follows: f_x = (e^(2x))_x * ln(x/y^2) + e^(2x) * (ln(x/y^2))_x. The derivative of e^(2x) is 2e^(2x), and the derivative of ln(x/y^2) simplifies to 1/x. Thus, the final expression for the derivative is f_x = 2e^(2x) * ln(x/y^2) + e^(2x) * (1/x).

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f(x,y)=e2x*ln(x/y^2)

find f(subscript-x)

Can anyone help me with this differentiation please? I know you must treat y as a constant and differentiate with respect to x.
 
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Just do it step by step. Since it is a product of two functions, use the product rule:
[itex]f_x= (e^{2x})_x (ln(x/y^2))+ (e^{2x})(ln(x/y^2))_x[/itex]

Now do [itex](e^{2x})_x[/itex] alone: the derivative of [itex]e^x[/itex] is just [itex]e^x[/itex] so by the chain rule the derivative is [itex]e^{2x}(2x)_x= 2e^{2x}[/itex].

Next do [itex](ln(x/y^2))_x[/itex] alone: the derivative of ln(x) is 1/x so by the chain rule the derivative is [itex](y^2/x)(x/y^2)_x= (y^2/x)(1/y^2)= 1/x[/itex].

Actually, that last could be done more easily: since [itex]ln(x/y^2)= ln(x)- 2ln(y)[/itex], the derivative is just 1/x.

Now put them all together.
 
Thanks very much I understand now
 

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