Question
Differentiate. \[ f(t)=\frac{2 t}{9+t^{2}} \] \( f^{\prime}(t)=\square \)
Ask by Mccoy Love. in the United States
Mar 15,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
\( f^{\prime}(t)=\frac{2(9-t^2)}{(9+t^2)^2} \)
Solution
1. Recall the quotient rule for differentiation: if
\[
f(t)=\frac{g(t)}{h(t)},
\]
then
\[
f'(t)=\frac{g'(t)h(t)-g(t)h'(t)}{[h(t)]^2}.
\]
2. Identify \( g(t) \) and \( h(t) \):
\[
g(t)=2t \quad \text{and} \quad h(t)=9+t^2.
\]
3. Differentiate \( g(t) \) and \( h(t) \):
\[
g'(t)=2,
\]
\[
h'(t)=2t.
\]
4. Substitute into the quotient rule:
\[
f'(t)=\frac{2(9+t^2)-2t(2t)}{(9+t^2)^2}.
\]
5. Simplify the numerator:
\[
2(9+t^2)=18+2t^2,
\]
and
\[
2t(2t)=4t^2.
\]
Thus,
\[
18+2t^2-4t^2 = 18-2t^2.
\]
6. Factor the common factor in the numerator:
\[
18-2t^2=2(9-t^2).
\]
7. The derivative is:
\[
f'(t)=\frac{2(9-t^2)}{(9+t^2)^2}.
\]
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Simplify this solution Mind Expander
To differentiate the function \( f(t) = \frac{2t}{9+t^2} \), we can use the quotient rule, which states that if \( f(t) = \frac{g(t)}{h(t)} \), then \( f'(t) = \frac{g'(t)h(t) - g(t)h'(t)}{(h(t))^2} \). Here, \( g(t) = 2t \) and \( h(t) = 9+t^2 \). Calculating the derivatives: - \( g'(t) = 2 \) - \( h'(t) = 2t \) Now, apply the quotient rule: \[ f'(t) = \frac{(2)(9+t^2) - (2t)(2t)}{(9+t^2)^2} = \frac{(18 + 2t^2 - 4t^2)}{(9+t^2)^2} = \frac{18 - 2t^2}{(9+t^2)^2} \] So, the final answer is: \[ f'(t) = \frac{18 - 2t^2}{(9+t^2)^2} \]
