Ladder rate-of-change problem
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why not do dy/dx multiplied by dx/dt ? im asking not saying thats the way to do it
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I agree. It really is a classic Rate of Change problem considering I just had a very similar problem during Math class today... :\
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Anyone else see the beat wearing an elf hat @4:08?
(just drawn)
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<3
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i love you sal!
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THANK YOU FOR HELPING ME PASS MY CALCULUS EXAM! =D
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it's the chain rule: since your differentiating with respect to a different variable, you find the derivative of the x^2 (2x), followed by the prime of x with respect to the other the variable (dx/dt). If confused, try taking the derivative of x^2 with respect to x (d/dx {x^2}), what you get is 2x*dx/dx (dx/dx is x prime) and the dx/dx cancel out to become 1. the value then become equal to 2x. Hope it helps
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at: 06:15 why do you have to multiply the derivative of x^2 which is 2x by dx/dy i thought it would be just 2x + 2(dy/dx) please help!!!
9:50Calculus: Derivatives 3by khanacademy181,727 views
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that kicked ass, thank you
jomorrissey07 2 years ago 18
Thank you for doing this and explaining it so well.
elfangor3 3 years ago 12