How do you find the derivative of (t^1.7 + 8)/(t^1.4 + 6)t1.7+8t1.4+6?

1 Answer
May 14, 2015

Quocient rule!

Let's just remember the quocient rule by definition, here:

be y=(f(x))/(g(x))y=f(x)g(x), then its derivative is given by:

y'=(f'(x)*g(x)-f(x)*g'(x))/(f(x)^2)y'=f'(x)g(x)f(x)g'(x)f(x)2.

Now, let's just derivate your function accordingly:

(dy)/(dx) = (1.7t^(0.7)*(t^(1.4)+6)-(t^1.7+8)(1.4t^0.4))/(t^1.7+8)^2dydx=1.7t0.7(t1.4+6)(t1.7+8)(1.4t0.4)(t1.7+8)2

(dy)/(dx) = ((1.7t^2.1+10.2t^0.7)-1.4t^2.1+11.2t^0.4)/(t^1.7+8)^2dydx=(1.7t2.1+10.2t0.7)1.4t2.1+11.2t0.4(t1.7+8)2

(dy)/(dx) = (0.3t^2.1+10.2t^0.7-11.2t^0.4)/(t^1.7+8)^2dydx=0.3t2.1+10.2t0.711.2t0.4(t1.7+8)2