Is this way correct
y=exp(k*x^2)
let u=k*x^2 => y=exp(u)=e^u
y'=dy/dx=(dy/du)*(du/dx)=(d(e^u)/du)*(du/dx)
=(e^u)*(2*k*x)
=2*k*x*exp(k*x^2)
(similar to 猫科动物's answer)
let u=k*x^2 => y=exp(u)=e^u
y'=dy/dx=(dy/du)*(du/dx)=(d(e^u)/du)*(du/dx)
=(e^u)*(2*k*x)
=2*k*x*exp(k*x^2)
(similar to 猫科动物's answer)
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