We have `y = e^(ln(x^2)(1+e^(-x)))`

To find the first derivative, we differentiate the function once.

To differentiate, use the chain rule: `d/(dx) f(g(x)) = f'(g(x))g'(x)`

Let `f(x) = e^x` and `g(x) = ln(x^2)(1+e^-x) = 2ln(x)(1+e^-x)`

(using the rule `ln(a^b) = bln(a)` )

Then `d/(dx) f(g(x)) = e^(ln(x^2)(1+e^(-x)))g'(x)` (because `f'(x)...

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We have `y = e^(ln(x^2)(1+e^(-x)))`

To find the first derivative, we differentiate the function once.

To differentiate, use the chain rule: `d/(dx) f(g(x)) = f'(g(x))g'(x)`

Let `f(x) = e^x` and `g(x) = ln(x^2)(1+e^-x) = 2ln(x)(1+e^-x)`

(using the rule `ln(a^b) = bln(a)` )

Then `d/(dx) f(g(x)) = e^(ln(x^2)(1+e^(-x)))g'(x)` (because `f'(x) = e^x`)

` `Now, `g'(x) = d/(dx) (2ln(x))(1+e^(-x)) + 2ln(x)d/(dx)(1+e^(-x))`

(using the multiplication rule of differentiation)

Which gives `g'(x) = 2/x(1+e^(-x)) + 2ln(x)(-e^(-x)) = 2/x(1+e^(-x)) -2ln(x)e^(-x)`

Plugging this back in to `d/(dx) f(g(x)) = f'(g(x))g'(x)`

Gives `d/(dx) f(g(x)) = e^(ln(x^2)(1+e^(-x)))(2/x(1+e^(-x))-2ln(x)e^(-x))`

**This is the first derivative dy/dx**