if $f(x)=(4x^{2}-5x + 6)csc(x)$, find $f(x)=$

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y'=u'v + uv'$. Let $u = 4x^{2}-5x + 6$ and $v=\csc(x)$. First, find $u'$ and $v'$. $u'=\frac{d}{dx}(4x^{2}-5x + 6)=8x - 5$ $v'=\frac{d}{dx}(\csc(x))=-\csc(x)\cot(x)$

Step2: Calculate $f'(x)$

Using the product - rule $f'(x)=u'v+uv'$, we substitute $u$, $u'$, $v$, and $v'$: $f'(x)=(8x - 5)\csc(x)+(4x^{2}-5x + 6)(-\csc(x)\cot(x))$ $f'(x)=(8x - 5)\csc(x)-(4x^{2}-5x + 6)\csc(x)\cot(x)$ $f'(x)=\csc(x)[8x - 5-(4x^{2}-5x + 6)\cot(x)]$ $f'(x)=\csc(x)(8x - 5-(4x^{2}\cot(x)-5x\cot(x)+6\cot(x)))$ $f'(x)=\csc(x)(8x - 5 - 4x^{2}\cot(x)+5x\cot(x)-6\cot(x))$

Answer:

$(8x - 5)\csc(x)-(4x^{2}-5x + 6)\csc(x)\cot(x)$