![]() So, this is going to be the derivative with respect to x of But this does not look like that, and the key is toĪppreciate that one over x is the same thing as ![]() The n with respect to x, so if we're taking the derivative of that, that that's going to be equal to, we take the exponent, bring it out front, and we've proven it in other videos, but this is gonna be n times x to the, and then we decrement the exponent. So at first, you might say, "How does the power rule apply here?" The power rule, just to remind ourselves, it tells us that if we're taking the derivative of x to What is that going to be equal to? Pause this video and try to figure it out. So let's say we take the derivative with respect to x of one over x. Going to do in this video is get some practice takingĭerivatives with the power rule. So, I think something doesn't work when there's a zero power in the denominator. So, for the KA problems, the formula you need isĮDIT 3: made a great point, saying that one of my calculations above, d/dx 1/x^0 = d/dx x^-0 = 0x^-1 = 1/( 0x^1 ) must be wrong, because 1/(0x^1) = 1/0 = Infinity is not equal to the common sense solution, d/dx 1/x^0 = d/dx x^-0 = d/dx 1 = 0. My answer actually WAS CORRECT, but it wasn't in the form they accept. I put in, according to my newly-acquired formula, Thanks for your time, you who have read down to here.Īlso, please tell me if some of my calculations are wrong and I'll correct them, I'm regularly online.ĭ/dx 1/x^a = d/dx x^-a = -ax^(-a-1) = 1/-ax^(a+1)ĮDIT 2: Do not try this on KA. Is this just something completely obvious and natural or is it really as awesome as I think?ģ. I feel like this is a true pattern, but can someone tell me:Ģ. I believe I've found an "Inverse power rule", where
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