How do you find the derivative of #(t^1.7 + 8)/(t^1.4 + 6)#?

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Answer 1 · 2015-05-14T20:27:31

Quocient rule!

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

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

#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)^2#

#(dy)/(dx) = ((1.7t^2.1+10.2t^0.7)-1.4t^2.1+11.2t^0.4)/(t^1.7+8)^2#

#(dy)/(dx) = (0.3t^2.1+10.2t^0.7-11.2t^0.4)/(t^1.7+8)^2#