Question

find the derivative of the function.
y = x sin(8/x)
y(x) =

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = uv$, where $u$ and $v$ are functions of $x$, then $y'=u'v + uv'$. Here, $u = x$ and $v=\sin(\frac{8}{x})$. The derivative of $u$ with respect to $x$ is $u'=\frac{d}{dx}(x)=1$.

Step2: Apply chain - rule to find $v'$

Let $t=\frac{8}{x}=8x^{-1}$. Then $v = \sin(t)$. By the chain - rule $\frac{dv}{dx}=\frac{dv}{dt}\cdot\frac{dt}{dx}$. The derivative of $\sin(t)$ with respect to $t$ is $\cos(t)$, and the derivative of $t = 8x^{-1}$ with respect to $x$ is $\frac{dt}{dx}=- 8x^{-2}=-\frac{8}{x^{2}}$. So $v'=\cos(\frac{8}{x})\cdot(-\frac{8}{x^{2}})=-\frac{8}{x^{2}}\cos(\frac{8}{x})$.

Step3: Calculate $y'$

Using the product - rule $y'=u'v + uv'$, we substitute $u = x$, $u' = 1$, $v=\sin(\frac{8}{x})$, and $v'=-\frac{8}{x^{2}}\cos(\frac{8}{x})$. So $y'=1\cdot\sin(\frac{8}{x})+x\cdot(-\frac{8}{x^{2}}\cos(\frac{8}{x}))=\sin(\frac{8}{x})-\frac{8}{x}\cos(\frac{8}{x})$.

Answer:

$\sin(\frac{8}{x})-\frac{8}{x}\cos(\frac{8}{x})$