`y = e^(ln(x^2)(1+e^(-x)))` What is the first derivative? The main problem is I have never dealt with In^2 previously
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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
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