Question

derivative:

1. f(t)= (t/t-3)^2 2. h(t) = (t^2/t^3+2)^2

Answer 1

f(t)= (t/t-3)^2
To find the derivative, apply the power rule, chain rule and quotient rule.

Answer 2

f(t) = (t)^2/(t-3)^2 ???then?

Answer 3

Use the quotient rule.

Answer 4

But I'd use the power rule first and then the quotient rule.
\[
f(t) = \left(\frac{t}{t-3}\right)^2 \\
f'(t) = 2\left(\frac{t}{t-3}\right)^1\frac{d}{dt} \left(\frac{t}{t-3}\right)
\]
Now apply the quotient rule.

Answer 5

the derivative of (t/t-3) is 0 right?

Answer 6

\[
\frac{d}{dt}\left(\frac{u}{v}\right) = \frac{vu'-uv'}{v^2} \\
\frac{d}{dt}\frac{t}{t-3} = \frac{(t-3)(1) - t(1)}{(t-3)^2} = \frac{-3}{(t-3)^2}
\]

Answer 7

Read about quotient rule in your book if it is not clear.

Answer 8

f'(t)= 2(t/t-3)x -3/(t-3)^2 then?>

Answer 9

\[
f'(t) = \frac{2t}{(t-3)} * \frac{-3}{(t-3)^2} = -\frac{-6t}{(t-3)^3}
\]

Answer 10

can you help me with others?

Answer 11

\[g(t)=\sqrt{\sqrt{t+1}} +1 \]