Question

If (y)sin(2y) = (x)cos(2x), then what is dy/dx at the point (pi/ 4, pi / 2)? -1/21/2-1/41/4

Answer 1

Given equation is

`(y)\sin (2y)=(x)(\cos (2x)`

Differentiate with respect to x, we get

`\text{ Using }(d(f(x)g(x)))/(dx)=f(x)(d(g(x))/(dx)+g(x)(df(x))/(dx)`

`(d(y))/(dx)\sin (2y)+y(d(\sin (2y)))/(dx)=(d(x))/(dx)\cos (2x)+x(d(\cos(2x)))/(dx)`

`(dy)/(dx)\sin (2y)+y\cos (2y)*2*(dy)/(dx)=\cos (2x)+x(-\sin (2x))*2`

`(dy)/(dx)(\sin (2y)+2y\cos (2y))=\cos (2x)-2x\sin (2x)`
`\text{ Substitute x=}(\pi)/(4)\text{ and y=}(\pi)/(2)\text{, we get}`

`(dy)/(dx)(\sin (2(\pi)/(2))+2(\pi)/(2)\cos (2(\pi)/(2)))=\cos (2(\pi)/(4))-2(\pi)/(4)\sin (2(\pi)/(4))`

`(dy)/(dx)(\sin (\pi)+\pi\cos (\pi))=\cos ((\pi)/(2))-(\pi)/(2)\sin ((\pi)/(2))`

`\text{ Use }\sin \pi=0,\cos \pi=-1,\cos ((\pi)/(2))=0,\text{ and }\sin ((\pi)/(2))=1\text{, we get}`

`(dy)/(dx)(0+\pi(-1))=0-(\pi)/(2)(1)`

`(dy)/(dx)(-\pi)=-(\pi)/(2)`

`(dy)/(dx)=-(\pi)/(2)*(1)/(-\pi)`

`(dy)/(dx)=(1)/(2)`

Hence the answer is

`(dy)/(dx)=(1)/(2)`