Why is the f '(x) = 2xe^(x^2) the derivative of f(x) =e^(2x)?
Apply the chain rule:
df / dx = df / du ∙ du / dx
where:
f = eᵘ , u = 2x
d ( eᵘ ) / du ∙ d ( 2 x ) / dx = eᵘ ∙ 2 = 2 eᵘ
substitute back u = 2 x
df / dx = 2 ∙ e²ˣ = 2 ∙ (eˣ)²
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Remark:
d ( eᵘ ) / du = eᵘ
d ( 2 x ) / dx = 2
eⁿˣ =(eˣ)ⁿ
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