Asked by Jennifer
Compute the indicated derivative.
d/dt [ ( t^2 + t^0.5 ) ( t^0.5 - t^-0.5 ) ] |t=1
I took the f and g prime then I used the quotient rule my final answer came out to
3t^1.5 - 1.5t^0.5 + t + 0.5t^-1
then I plugged 1 in for t and got 5.196
Please let me know if this is right, if not where did I go wrong.
thank you so much.
d/dt [ ( t^2 + t^0.5 ) ( t^0.5 - t^-0.5 ) ] |t=1
I took the f and g prime then I used the quotient rule my final answer came out to
3t^1.5 - 1.5t^0.5 + t + 0.5t^-1
then I plugged 1 in for t and got 5.196
Please let me know if this is right, if not where did I go wrong.
thank you so much.
Answers
Answered by
Steve
It appears to me that you have a product, not a quotient.
If
f(x) = (t^2 + t^0.5)(t^0.5 - t^-0.5)
f'(x) = (2t+0.5t^-0.5)(0.5t^-0.5 + 0.5t^-1.5)
= 2.5t^1.5 - 1.5t^0.5 + 1
This can be checked since if we expand f(x), we get
f(x) = t^2.5 - t^1.5 + t - 1
f'(1) = 2.5-1.5+1 = 2
Even if your f' were correct, I have no idea where all those extra decimal places came from. 1^n = 1 for any power n.
If you did indeed have a quotient,
f(t) = (t^2 + t^0.5)/(t^0.5 - t^-0.5)
f(1) is undefined.
If
f(x) = (t^2 + t^0.5)(t^0.5 - t^-0.5)
f'(x) = (2t+0.5t^-0.5)(0.5t^-0.5 + 0.5t^-1.5)
= 2.5t^1.5 - 1.5t^0.5 + 1
This can be checked since if we expand f(x), we get
f(x) = t^2.5 - t^1.5 + t - 1
f'(1) = 2.5-1.5+1 = 2
Even if your f' were correct, I have no idea where all those extra decimal places came from. 1^n = 1 for any power n.
If you did indeed have a quotient,
f(t) = (t^2 + t^0.5)/(t^0.5 - t^-0.5)
f(1) is undefined.
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