find the derivative of the trigonometric function.\n f(x)=\frac{sin(x)}{x^{6}} \n f(x)=square

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = \sin(x)$ and $v=x^{6}$.

Step2: Find $u'$ and $v'$

The derivative of $u=\sin(x)$ is $u'=\cos(x)$, and the derivative of $v = x^{6}$ using the power - rule $(x^{n})'=nx^{n - 1}$ is $v'=6x^{5}$.

Step3: Apply quotient - rule

Substitute $u$, $u'$, $v$, and $v'$ into the quotient - rule formula: [ \begin{align*} f'(x)&=\frac{\cos(x)\cdot x^{6}-\sin(x)\cdot6x^{5}}{(x^{6})^{2}}\ &=\frac{x^{5}(x\cos(x)-6\sin(x))}{x^{12}}\ &=\frac{x\cos(x)-6\sin(x)}{x^{7}} \end{align*} ]

Answer:

$\frac{x\cos(x)-6\sin(x)}{x^{7}}$