Find derivative of a trig function.

Question

If $y=\csc(xy)$, then find $y'$ in respect to $x$

$y'=-\csc(xy)\cdot \cot(xy)\cdot(xy)'$

$(xy)'=(x)'(y)+(x)(y)'$

$(xy)'=y+xy'$

$y'=-(y+xy')[\csc(xy)\cdot\cot(x)]$

AND, that's where I get lost. So I expanded.

$y'=-y[\csc(xy)\cdot\cot(xy)]-xy'[\csc(xy)\cdot\cot(xy)]$

$y'+xy'[\csc(xy)\cdot\cot(xy)]=-y[\csc(xy)\cdot\cot(xy)]$

$y'[(1+x)[\csc(xy)\cdot\cot(xy)]=\dfrac{-y[\csc(xy)\cdot\cot(xy)]}{ [(1+x)[\csc(xy)\cdot\cot(xy)]}$

$y'=\dfrac{-y}{1+x}$

Is this even right?

Answer 1

No. You've got a small mistake in your factoring near the end of your work.

$y'+xy'[\csc(xy)\cdot\cot(xy)]=-y[\csc(xy)\cdot\cot(xy)]\ \color{red}{\mathrm{correct\ to\ here}}\\ y'\left(1+x[\csc(xy)\cdot\cot(xy)]\right)=-y\csc(xy)\cdot\cot(xy) \\ y'=\dfrac{-y\csc(xy)\cdot\cot(xy)}{\left(1+x\csc(xy)\cdot\cot(xy)\right)} $