determine $g(t)$ if $g(t)=\frac{sin(t)sqrt{5t}}{(13 - t^{2}+6t^{3})^{8}}$. $g(t)=$

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $g(t)=\frac{u(t)}{v(t)}$, then $g^{\prime}(t)=\frac{u^{\prime}(t)v(t)-u(t)v^{\prime}(t)}{v(t)^2}$. Here, $u(t)=\sin(t)\sqrt{5t}=\sqrt{5}t^{\frac{1}{2}}\sin(t)$ and $v(t)=(13 - t^{2}+6t^{3})^{8}$.

Step2: Find $u^{\prime}(t)$

Use the product - rule $(uv)^\prime = u^\prime v+uv^\prime$. Let $a = \sqrt{5}t^{\frac{1}{2}}$ and $b=\sin(t)$. Then $a^\prime=\frac{\sqrt{5}}{2}t^{-\frac{1}{2}}$ and $b^\prime=\cos(t)$. So $u^{\prime}(t)=\sqrt{5}\left(\frac{1}{2}t^{-\frac{1}{2}}\sin(t)+t^{\frac{1}{2}}\cos(t)\right)$.

Step3: Find $v^{\prime}(t)$

Use the chain - rule. If $y = f(g(x))$, then $y^\prime=f^\prime(g(x))g^\prime(x)$. Let $y = u^{8}$ and $u = 13 - t^{2}+6t^{3}$. Then $y^\prime = 8u^{7}$ and $u^\prime=-2t + 18t^{2}$. So $v^{\prime}(t)=8(13 - t^{2}+6t^{3})^{7}(-2t + 18t^{2})$.

Step4: Calculate $g^{\prime}(t)$

$g^{\prime}(t)=\frac{\sqrt{5}\left(\frac{1}{2}t^{-\frac{1}{2}}\sin(t)+t^{\frac{1}{2}}\cos(t)\right)(13 - t^{2}+6t^{3})^{8}-\sin(t)\sqrt{5t}\times8(13 - t^{2}+6t^{3})^{7}(-2t + 18t^{2})}{(13 - t^{2}+6t^{3})^{16}}$ Simplify the above expression: [ \begin{align*} g^{\prime}(t)&=\frac{\sqrt{5}(13 - t^{2}+6t^{3})^{7}\left[\left(\frac{1}{2}t^{-\frac{1}{2}}\sin(t)+t^{\frac{1}{2}}\cos(t)\right)(13 - t^{2}+6t^{3})-8\sin(t)\sqrt{5t}(-2t + 18t^{2})\right]}{(13 - t^{2}+6t^{3})^{16}}\ &=\frac{\sqrt{5}\left[\frac{1}{2}t^{-\frac{1}{2}}\sin(t)(13 - t^{2}+6t^{3})+t^{\frac{1}{2}}\cos(t)(13 - t^{2}+6t^{3})-8\sin(t)\sqrt{5t}(-2t + 18t^{2})\right]}{(13 - t^{2}+6t^{3})^{9}} \end{align*} ]

Answer:

$\frac{\sqrt{5}\left[\frac{1}{2}t^{-\frac{1}{2}}\sin(t)(13 - t^{2}+6t^{3})+t^{\frac{1}{2}}\cos(t)(13 - t^{2}+6t^{3})-8\sin(t)\sqrt{5t}(-2t + 18t^{2})\right]}{(13 - t^{2}+6t^{3})^{9}}$