Question

Find dy/dx by implicit differentiation for ysin(y) = xcos(x)

Answer 1

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

Explanation:

So we have:

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

And we want to find dy/dx.

So, let's take the derivative of both sides with respect to x:

`(d)/(dx)[y\sin(y)]=(d)/(dx)[x\cos(x)]`

Let's do each side individually.

Left Side:

We have:

`(d)/(dx)[y\sin(y)]`

We can use the product rule:

`(uv)'=u'v+uv'`

So, our derivative is:

`=(d)/(dx)[y]\sin(y)+y(d)/(dx)[\sin(y)]`

We must implicitly differentiate for y. This gives us:

`=(dy)/(dx)\sin(y)+y(d)/(dx)[\sin(y)]`

For the sin(y), we need to use the chain rule:

`u(v(x))'=u'(v(x))\cdot v'(x)`

Our u(x) is sin(x) and our v(x) is y. So, u'(x) is cos(x) and v'(x) is dy/dx.

So, our derivative is:

`=(dy)/(dx)\sin(y)+y(\cos(y)\cdot(dy)/(dx)})`

Simplify:

`=(dy)/(dx)\sin(y)+y\cos(y)\cdot(dy)/(dx)}`

And we are done for the right.

Right Side:

We have:

`(d)/(dx)[x\cos(x)]`

This will be significantly easier since it's just x like normal.

Again, let's use the product rule:

`=(d)/(dx)[x]\cos(x)+x(d)/(dx)[\cos(x)]`

Differentiate:

`=\cos(x)-x\sin(x)`

So, our entire equation is:

`=(dy)/(dx)\sin(y)+y\cos(y)\cdot(dy)/(dx)}=\cos(x)-x\sin(x)`

To find our derivative, we need to solve for dy/dx. So, let's factor out a dy/dx from the left. This yields:

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

Finally, divide everything by the expression inside the parentheses to obtain our derivative:

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

And we're done!