Question

Let x^4+y^4=16 and consider y as a function of x. use the implicit differentiation to find y"

Answer 1

`(dy)/(dx) = - \frac{ {x}^(3) }{ {y}^(3) }`

Explanation:

Differentiate both sides of the equation (consider y as a function of x).

`(d)/(dx) ( {x}^(4) + {y}^(4) (x)) = (d)/(dx) (16)`

the derivative of a sum/difference is the sum/difference of derivatives.

`( (d)/(dx) ( {x}^(4) + {y}^(4) (x))`

`= ( (d)/(dx) ( {x}^(4) ) + (d)/(dx) ( {y}^(4) (x)))`

the function of y^4(x) is the composition of f(g(x)) of the two functions.

the chain rule:

`(d)/(dx) (f(g(x))) = (d)/(du) (f(u)) (d)/(dx) (g(x))`

`= ( (d)/(du) ( {u}^(4) ) (d)/(dx) (y(x))) + (d)/(dx) ( {x}^(4) )`

apply the power rule:

`(4 {u}^(3) ) (d)/(dx) (y(x)) + (d)/(dx) ( {x}^(4) )`

return to the old variable:

`4(y(x) {)}^(3) (d)/(dx) (y(x)) + (d)/(dx) ( {x}^(4) )`

apply the power rule once again:

`4 {y}^(3) (x) (d)/(dx) (y(x)) + (4 {x}^(3) )`

simplify:

`4 {x}^(3) + 4 {y}^(3) (x) (d)/(dx) (y(x))`

`= 4( {x}^(3) + {y}^(3) (x) (d)/(dx) (y(x)))`

`= (d)/(dx) ( {x}^(4) + {y}^(4) ))`

`= 4( {x}^(3) + {y}^(3) (x) (d)/(dx)(y(x)))`

differentiate the equation:

`( (d)/(dx) (16)) = (0)`

`= (d)/(dx) (16) = 0`

derivative:

`4 {x}^(3) + 4 {y}^(3) (dy)/(dx) = 0`

`(dy)/(dx) = - \frac{ {x}^(3) }{ {y}^(3) }`