Question

Find y=sin(x cot(2x-1)). Do not simplify the result. dy dx

Answer 1

`(dy)/(dx)=-2x\csc^2(2x-1)\cos(x\cot(2x-1))+\cot(2x-1)\cos(x\cot(2x-1))`

Explanation:

Differentiate the given expression.

`y=\sin(x \cot(2x-1))`

I will be using the following rules of differentiation...

`\boxed{\left\begin{array}{ccc}\text{\underline{The Chain Rule:}}\\\\(d)/(dx)[f(g(x))]=f'(g(x)) \cdot g'(x) \end{array}\right}\\\\ \\\boxed{\left\begin{array}{ccc}\text{\underline{The Product Rule:}}\\\\(d)/(dx)[f(x)g(x)]=f(x)g'(x)+f'(x)g(x) \end{array}\right} \\ \\ \\ \boxed{\left\begin{array}{ccc}\text{\underline{Sine Rule:}}\\\\(d)/(dx)[\sin(x)]=\cos(x) \end{array}\right}`

`\boxed{\left\begin{array}{ccc}\text{\underline{Cotanget Rule:}}\\\\(d)/(dx)[\cot(x)]=-\csc^2(x) \end{array}\right} \\ \\ \\ \boxed{\left\begin{array}{ccc}\text{\underline{Power Rule:}}\\\\(d)/(dx)[x^n]=nx^(n-1) \end{array}\right}`

`\hrulefill`

`y=\sin(x \cot(2x-1))\\\\\Longrightarrow (dy)/(dx) =(d)/(dx)[\sin()] \\\\\Longrightarrow (dy)/(dx) =\cos(x \cot(2x-1))\cdot(d)/(dx)[x \cot(2x-1)] \\\\\Longrightarrow (dy)/(dx) =\cos(x \cot(2x-1))\cdot[x (d)/(dx)[\cot(2x-1)]+(d)/(dx)[x]\cot(2x-1)] \\\\\Longrightarrow (dy)/(dx) =\cos(x \cot(2x-1))\cdot[-x \csc^2(2x-1)\cdot(d)/(dx)[2x+1] +\cot(2x-1)] \\\\\Longrightarrow (dy)/(dx) =\cos(x \cot(2x-1))\cdot[-2x \csc^2(2x-1)+\cot(2x-1)] \\\\`

`\therefore \boxed{(dy)/(dx)=-2x\csc^2(2x-1)\cos(x\cot(2x-1))+\cot(2x-1)\cos(x\cot(2x-1)) }`