Differentiate y=(t^3 + 3t)/ (t^2-4t+3)

Question

Please help me 

Answer 1

Hi! Let's think of the two functions above and below the central line as P(t) and Q(t). The Quotient Rule has that if P and Q are functions of t, and their derivatives in t are P' and Q', then the derivative of y is: dydt = P'Q - PQ' -------- Q^2 So, here's the differentiation of both functions: P(t) = t^3 + 3t P'(t) = 3t^2 + 3 Q(t) = t^2 - 4t + 3 Q'(t) = 2t - 4 So, the derivative of y is: dydt = [3t^2 + 3][t^2 - 4t + 3] - [t^3 + 3t][2t - 4] --------------------------------------------- [t^2 - 4t + 3]^2 Here's the algebra afterward, so you can check yourself: [3t^2 + 3][t^2 - 4t + 3] = 3t^4 - 12t^3 + 9t^2 + 3t^2 - 12t + 9 ----------------------------- 3t^4 - 12t^3 + 12t^2 - 12t + 9 [t^3 + 3t][2t - 4] = 2t^4 - 4t^3 + 6t^2 - 12t ------------------------ 2t^4 - 4t^3 + 6t^2 - 12t [t^2 - 4t + 3]^2 = t^4 - 4t^3 + 3t^2 - 4t^3 + 16t^2 - 12t 3t^2 - 12t + 9 ---------------------------- t^4 - 8t^3 + 22t^2 - 24t + 9 dydt = 3t^4 - 12t^3 + 12t^2 - 12t + 9 - [2t^4 - 4t^3 + 6t^2 - 12t] ----------------------------------------------------------- t^4 - 8t^3 + 22t^2 - 24t + 9 = t^4 - 8t^3 + 6t^2 + 9 ---------------------------- t^4 - 8t^3 + 22t^2 - 24t + 9

Answer 2

Jessica, 

 

Here you use the quotient rule:

dy/dx (u/v)=(vu'-uv')/v^2

Let u=t^3+3t and v=t^2-4t+3

Then u'=3t^2+3 and v'=2t-4

 

Now plugging into the formula:

[(t^2-4t+3)(3t^2+3)-(t^3+3t)(2t-4)]/(t^2-4t+3)^2

 

Things get fairly nasty from here so expand carefully!