Question

differentiation of x^4+y^4=0​

Answer 1

Answer:

4x^3 + 4 y^3 × dy/dx =0

dy/dx = - x^3 / y^3

Answer 2

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

Given:

• ${x}^{4} + {y}^{4} = 0$

To find:

• Differentiate the equation.

Solution:

Formula to be used:

1. $\bf \frac{d}{dx} ( {y}^{n} ) = n {y}^{n - 1} \frac{dy}{dx}$ and
2. $\bf \frac{d}{dx} ( {x}^{n} ) = n {x}^{n - 1} \frac{dx}{dx} = n {x}^{n - 1}$

Step 1:

Apply the formula for differentiation.

$\frac{d}{dx} ( {x}^{4} + {y}^{4} ) = \frac{d}{dx} (0)$

so,

$4 {x}^{4 - 1} + 4 {y}^{4 - 1} \frac{dy}{dx} = 0$

or

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

Step 2:

Cancel the common terms and simplify.

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

or

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

Thus,

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

Learn more:

1) Differentiate the identity sin3x=3sinx—4sin³x. What do you observe ?

https://brainly.in/question/11771928

2) if y=x^x, then dy/dx is equal to

https://brainly.in/question/17124124