hw12 derivatives of trigonometric functions (target c3; $3.5)\nscore: 2/7 answered: 2/7\nquestion 3\nlet…

Explanation:

Step1: Find derivative of u(x)

The derivative of $\sin(x)$ is $\cos(x)$, so $u'(x)=\cos(x)$.

Step2: Find derivative of v(x)

Using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, for $v(x)=x^{18}$, $v'(x)=18x^{17}$.

Step3: Find derivative of f(x)

We know $f(x)=\frac{u(x)}{v(x)}$ and $f'(x)=\frac{u'v - uv'}{v^{2}}$. Substitute $u(x)=\sin(x)$, $u'(x)=\cos(x)$, $v(x)=x^{18}$, and $v'(x)=18x^{17}$ into the quotient - rule formula. $f'(x)=\frac{\cos(x)\cdot x^{18}-\sin(x)\cdot18x^{17}}{(x^{18})^{2}}=\frac{x^{17}(x\cos(x)-18\sin(x))}{x^{36}}=\frac{x\cos(x)-18\sin(x)}{x^{19}}$.

Answer:

$u'(x)=\cos(x)$ $v'(x)=18x^{17}$ $f'(x)=\frac{x\cos(x)-18\sin(x)}{x^{19}}$