Question

find the derivative of the given function.
q = sin(\frac{t}{\sqrt{t + 5}})
\frac{dq}{dt}=\square

Explanation:

Step1: Apply chain - rule

Let $u=\frac{t}{\sqrt{t + 5}}$, then $q=\sin(u)$. The chain - rule states that $\frac{dq}{dt}=\frac{dq}{du}\cdot\frac{du}{dt}$. First, find $\frac{dq}{du}$. Since $q = \sin(u)$, $\frac{dq}{du}=\cos(u)=\cos(\frac{t}{\sqrt{t + 5}})$.

Step2: Find $\frac{du}{dt}$ using quotient - rule

The quotient - rule for $u=\frac{f(t)}{g(t)}$ where $f(t)=t$ and $g(t)=\sqrt{t + 5}=(t + 5)^{\frac{1}{2}}$ is $\frac{du}{dt}=\frac{f^{\prime}(t)g(t)-f(t)g^{\prime}(t)}{g^{2}(t)}$. We know that $f^{\prime}(t)=1$ and $g^{\prime}(t)=\frac{1}{2}(t + 5)^{-\frac{1}{2}}$. Then $\frac{du}{dt}=\frac{1\cdot\sqrt{t + 5}-t\cdot\frac{1}{2}(t + 5)^{-\frac{1}{2}}}{t + 5}=\frac{\sqrt{t + 5}-\frac{t}{2\sqrt{t + 5}}}{t + 5}=\frac{\frac{2(t + 5)-t}{2\sqrt{t + 5}}}{t + 5}=\frac{2t+10 - t}{2(t + 5)^{\frac{3}{2}}}=\frac{t + 10}{2(t + 5)^{\frac{3}{2}}}$.

Step3: Calculate $\frac{dq}{dt}$

By the chain - rule $\frac{dq}{dt}=\frac{dq}{du}\cdot\frac{du}{dt}=\cos(\frac{t}{\sqrt{t + 5}})\cdot\frac{t + 10}{2(t + 5)^{\frac{3}{2}}}$.

Answer:

$\frac{t + 10}{2(t + 5)^{\frac{3}{2}}}\cos(\frac{t}{\sqrt{t + 5}})$