Question

find the derivative of the trigonometric function.
g(v)=\frac{cos(v)}{csc(v)}
g(v)= square

Explanation:

Step1: Simplify the function

Recall that $\csc(v)=\frac{1}{\sin(v)}$, so $g(v)=\frac{\cos(v)}{\csc(v)}=\cos(v)\sin(v)$.

Step2: Use product - rule for differentiation

The product - rule states that if $y = u\cdot v$, then $y^\prime=u^\prime v + uv^\prime$. Let $u = \cos(v)$ and $v=\sin(v)$. Then $u^\prime=-\sin(v)$ and $v^\prime=\cos(v)$.
$g^\prime(v)=-\sin(v)\sin(v)+\cos(v)\cos(v)$.

Step3: Simplify the result

Using the identity $\cos^{2}(v)-\sin^{2}(v)=\cos(2v)$, we have $g^\prime(v)=\cos(2v)$.

Answer:

$\cos(2v)$