Question

How do you differentiate `sinx+xy+y^5=pi`?

Answer 1

`dy/dx=-(y+cosx)/(x+5y^4)`

Explanation:

`sinx +xy +y^5=pi`
Differentiating both sides with respect to `x` we get
`cosx+(y+xdy/dx)+5y^4dy/dx=0`
`:.dy/dx=-(y+cosx)/(x+5y^4)`

Answer 2

Given: `sinx+xy+y^5=pi`

Differentiate each term:

`(d(sinx))/dx+(d(xy))/dx+(d(y^5))/dx=(d(pi))/dx`

The first term is well known `(d(sinx))/dx = cos(x)`:

`cos(x)+(d(xy))/dx+(d(y^5))/dx=(d(pi))/dx`

The second term requires the use of the product rule:

`(d(xy))/dx = (d(x))/dxy+x(d(y))/dx`

`(d(xy))/dx = y+xdy/dx`

`cos(x)+y+xdy/dx+(d(y^5))/dx=(d(pi))/dx`

The third term requires the use of the chain rule:

`(d(y^5))/dx = (d(y^5))/dydy/dx`

`(d(y^5))/dx = 5y^4dy/dx`

`cos(x)+y+xdy/dx+5y^4dy/dx=(d(pi))/dx`

For the last term, we invoke the fact that the derivative of a constant is 0:

`cos(x)+y+xdy/dx+5y^4dy/dx=0`

Move all of the terms NOT containing `dy/dx` to the right:

`xdy/dx+5y^4dy/dx=-(cos(x)+y)`

Factor out `dy/dx` on the left:

`(x+5y^4)dy/dx=-(cos(x)+y)`

Divide both sides by the leading coefficient:

`dy/dx=-(cos(x)+y)/(x+5y^4)`