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# g^((3))(x) = 2e^x e^x xe^x # # g^((3))(x) = 3e^x xe^x # So it looks like clear pattern is forming, but let us just check by looking at the fourth derivative;Solve your math problems using our free math solver with stepbystep solutions Our math solver supports basic math, prealgebra, algebra, trigonometry, calculus and moreSolve your math problems using our free math solver with stepbystep solutions Our math solver supports basic math, prealgebra, algebra, trigonometry, calculus and more
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If you solve y'=2y by separation, as y'/y=2, ln(y)=2xC, y=e2xC then along the way you've assumed y>0, and those are the solutions y=e C e2x can represent The true general solution is y=Ce2x, and often people hand wave at this point and pretend e C is the same as C, because it gets the right answer But it's not legit The better way to solve that equation is multiplying by theP X e C X g ʼn돮 i ܂ Ȃ j z J ߂ȃ^ b ` ŏ ̎q ̃ g g ` B f W ^ i R y C ^ g p j Ȃ̂Ńf ^ i \ B ₳ G B MARINE NAVI l ́u 点 v p B \ ݂ ƁA A ܂ Ȃ V G ǎ T C g ɍڂ Ƃ ɁA ł m 点 v ܂ B Ȃ ̃ A h X ͎ ɂ͓` Ȃ V X e ł ̂ŁA \ ݂͂ǂ C y ɁBMARINE NAVI l ́u ܂ ɂイ v z M ܂ B This code is calculating the factorial , e and e^x and takes the input from the user, the x value in e^x is the same value taken from the user i can't get the same result for e^x as MathExp for some reason although it calculates the factorial and e currently NOTE e^x formula is e^x = 1x/1!




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Quadratic Function
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