Question

How do you differentiate `x-cos(x^2)+y^2/x+3x^5=4x^3`?

Answer 1

`dy/dx = (y^2+12x^4 - 15x^6 -x^2 - 2x^3 sin(x^2))/(2xy)`

Explanation:

We have:

`x-cos(x^2)+y^2/x+3x^5=4x^3`

Method 1: Implicit differentiation, as is:

Differentiating wrt `x` whilst applying the chain rule and the product rule:

`1 + sin(x^2)(d/dx x^2) + ( (x)(d/dxy^2) - (y^2)(d/dx x) ) / (x)^2 + 15x^4 = 12x^2`

`:. 1 + sin(x^2)(2x) + ( (x)(2ydy/dx) - (y^2)(1) ) / (x)^2 + 15x^4 = 12x^2`

`:. 1 + 2x sin(x^2) + ( 2xydy/dx - y^2 ) / x^2 + 15x^4 = 12x^2`

`:. ( 2xydy/dx - y^2 ) / x^2 = 12x^2 - 15x^4 - 1 - 2x sin(x^2)`

`:. 2xydy/dx - y^2 = 12x^4 - 15x^6 -1 - 2x^3 sin(x^2)`

`:. 2xydy/dx = y^2+12x^4 - 15x^6 -x^2 - 2x^3 sin(x^2)`

`:. dy/dx = (y^2+12x^4 - 15x^6 -x^2 - 2x^3 sin(x^2))/(2xy)`

Method 2: Implicit Function Theorem

Putting:

`F(x,y) = x-cos(x^2)+y^2/x+3x^5-4x^3`

We have:

`(partial F)/(partial x) = 1 + 2xsin(x^2) - y^2/x^2 + 15x^4 - 12x^2`

`(partial F)/(partial y) = (2y)/x`

Then:

`dy/dx = - ((partial F)/(partial x)) / ((partial F)/(partial y))`
`\ \ \ \ \ \ = - (1 + 2xsin(x^2) - y^2/x^2 + 15x^4 - 12x^2) / ( (2y)/x )`
`\ \ \ \ \ \ = - ( (x) (1 + 2xsin(x^2) - y^2/x^2 + 15x^4 - 12x^2) ) / (2y)`

Yielding the same result as above.