Calculate the Derivative in Two Ways

Question

First use the Product or Quotient Rule. Then rewrite the function algebraically and directly calculate the derivative.

h(t)= (t^2-1)/(t-1)

Answer 1

h(t)= (t²-1)/(t-1)

Using the Quotient Rule:

The bottom times the derivative of the top minus the top times the derivative of the bottom, all over the bottom squared:

h'(t)=[(t-1)(2t)-(t²-1)(1)]/(t-1)²=[2t²-2t-t2+1]/(t-1)²=[t²-2t+1]/(t-1)²=[(t-1)²]/(t-1)²=1

Using the Product Rule:

The first time the derivative of the second plus the second times the derivative of the first:

h(t)=(t²-1)(t-1)^-1

h'(t)=(t²-1)(-1)(t-1)^-2 + (t-1)^-1(2t)=-[(t+1)(t-1)]/(t-1)²+2t/(t-1)

=-[(t+1)/(t-1)]+2t/(t-1)=[1/(t-1)][-(t+1)+2t]=[1/(t-1)][t-1]=1

Now simplify algebraically before taking the derivative:

h(t)=(t²-1)/(t-1)=[(t+1)(t-1)]/(t-1)=t+1

h'(t)=1

Lesson learned: always simplify algebraically to save time before finding the derivative.

Answer 2

Product Rule:

y = (t^2-1)*(t-1)^(-1)

y ' = (t^2 - 1) (-1) (t-1)^(-2) + (2t) (t-1)^(-1)

= (t-1)^(-2) (1 - t^2) + (2t)(t-1) )

= (t-1)^(-2) ( 1 - t^2 + 2t^2 - 2t)

= (t^2 - 2t + 1)/ (t-1)^2

= (t - 1)^2/ (t-1)^2

= 1

Quotient Rule:

((t-1) (2t) - (t^2 -1) )/(t-1)^2 =

(2t^2 - 2t - t^2 + 1)/(t-1)^2 =

(t^2-2t + 1)/(t-1)^2 =

(t-1)^2/(t-1)^2 =

1

Finally, note that

y = (t+1)(t-1)/(t-1) = t+1

then y' = 1