y(x)=(1-t)/(t+1)^3 y'(x)=(t+1)^3(-1)(1-t) - (1-t)(3)(t+1)^2/(t+1)^6 y'(x)=(t+1)^3(t-1) - (1-t)(3)(t+1)^2/(t+1)^6 y'(x)=(t+1)^2[3t^2(t+1)]/(t+1)^6 y'(x)=3t^2(t+1)/(t+1)^4
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Originally Posted by becky y(x)=(1-t)/(t+1)^3 It is not correct. When you take the derivative of, you have, -1 Not,
y(x)=(1-t)/(t+1)^3 y'(x) = (t+1)^3 (-1) - (1-t)(3)(t+1)^2/(t+1)^6 Y'(x) = -3(t+1)^2[(t+1)(1-t)]/(t+1)^6
Originally Posted by becky y(x)=(1-t)/(t+1)^3 y'(x) = (t+1)^3 (-1) - (1-t)(3)(t+1)^2/(t+1)^6 Correct Y'(x) = -3(t+1)^2[(t+1)(1-t)]/(t+1)^6 Wrong. (Factorization was wrong). ---Let Die Meister attemptum, Thus, Thus, Thus, Thus,
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