Question

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

Answer 1

Answer:

$\frac{dy}{dx} = - \frac{ {x}^{3} }{ {y}^{3} }$

Step-by-step explanation:

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

$\frac{d}{dx} ( {x}^{4} + {y}^{4} (x)) = \frac{d}{dx} (16)$

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

$( \frac{d}{dx} ( {x}^{4} + {y}^{4} (x))$

$= ( \frac{d}{dx} ( {x}^{4} ) + \frac{d}{dx} ( {y}^{4} (x)))$

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

the chain rule:

$\frac{d}{dx} (f(g(x))) = \frac{d}{du} (f(u)) \frac{d}{dx} (g(x))$

$= ( \frac{d}{du} ( {u}^{4} ) \frac{d}{dx} (y(x))) + \frac{d}{dx} ( {x}^{4} )$

apply the power rule:

$(4 {u}^{3} ) \frac{d}{dx} (y(x)) + \frac{d}{dx} ( {x}^{4} )$

return to the old variable:

$4(y(x) {)}^{3} \frac{d}{dx} (y(x)) + \frac{d}{dx} ( {x}^{4} )$

apply the power rule once again:

$4 {y}^{3} (x) \frac{d}{dx} (y(x)) + (4 {x}^{3} )$

simplify:

$4 {x}^{3} + 4 {y}^{3} (x) \frac{d}{dx} (y(x))$

$= 4( {x}^{3} + {y}^{3} (x) \frac{d}{dx} (y(x)))$

$= \frac{d}{dx} ( {x}^{4} + {y}^{4} ))$

$= 4( {x}^{3} + {y}^{3} (x) \frac{d}{dx}(y(x)))$

differentiate the equation:

$( \frac{d}{dx} (16)) = (0)$

$= \frac{d}{dx} (16) = 0$

derivative:

$4 {x}^{3} + 4 {y}^{3} \frac{dy}{dx} = 0$

$\frac{dy}{dx} = - \frac{ {x}^{3} }{ {y}^{3} }$