Question

find the derivative of the given function.

q = \tan\left(\frac{t}{\sqrt{t + 3}}\
ight)

dq/dt =

Explanation:

Step1: Let \(u = \frac{t}{\sqrt{t + 3}}\)

Then \(q=\tan(u)\). By the chain rule \(\frac{dq}{dt}=\frac{dq}{du}\cdot\frac{du}{dt}\).
Since \(\frac{d}{du}(\tan(u))=\sec^{2}(u)\), so \(\frac{dq}{du}=\sec^{2}(u)=\sec^{2}(\frac{t}{\sqrt{t + 3}})\).

Step2: Find \(\frac{du}{dt}\) using the quotient rule

The quotient rule states that if \(u=\frac{f(t)}{g(t)}\) (\(f(t)=t\), \(g(t)=\sqrt{t + 3}=(t + 3)^{\frac{1}{2}}\)), then \(\frac{du}{dt}=\frac{f^{\prime}(t)g(t)-f(t)g^{\prime}(t)}{g^{2}(t)}\).
\(f^{\prime}(t) = 1\), \(g^{\prime}(t)=\frac{1}{2}(t + 3)^{-\frac{1}{2}}\) (by the power - rule \((x^{n})^\prime=nx^{n - 1}\)).

$$
LATEXBLOCK0
$$

Step3: Combine using the chain rule

\(\frac{dq}{dt}=\frac{dq}{du}\cdot\frac{du}{dt}=\sec^{2}(\frac{t}{\sqrt{t + 3}})\cdot\frac{t + 6}{2(t + 3)^{\frac{3}{2}}}\)

Answer:

\(\frac{t + 6}{2(t + 3)^{\frac{3}{2}}}\sec^{2}(\frac{t}{\sqrt{t+3}})\)