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Parametric differentiation question

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For question 9c, I understand everything apart from the steps they did on the last step on the mark scheme.
Reply 1
Original post by Matheen1
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For question 9c, I understand everything apart from the steps they did on the last step on the mark scheme.

One way to think about it is the fraction is multiplied by 1 = e^(3t)/e^(3t) then the exponents add when you multiply the numeator and denominator each by e^(3t). They also flip the sign of the numerator and denominator as well (so really the multiplier was 1 = (-e^(3t))/(-e^(3t)))
Reply 2
Original post by mqb2766
One way to think about it is the fraction is multiplied by 1 = e^(3t)/e^(3t) then the exponents add when you multiply the numeator and denominator each by e^(3t). They also flip the sign of the numerator and denominator as well (so really the multiplier was 1 = (-e^(3t))/(-e^(3t)))

Thanks, it all makes sense now.

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