Is this way correcty=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)