Find the derivative of x^3y+y^3x=30

\frac{d}{dx}\left(x^3y+y^3x=30\right)

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Answer

$y^3+3yx^{2}=0$

Step by step solution

Problem

$\frac{d}{dx}\left(x^3y+y^3x=30\right)$
1

Apply implicit differentiation by taking the derivative of both sides of the equation with respect to the differentiation variable

$\frac{d}{dx}\left(xy^3+yx^3\right)=\frac{d}{dx}\left(30\right)$
2

The derivative of the constant function is equal to zero

$\frac{d}{dx}\left(xy^3+yx^3\right)=0$
3

The derivative of a sum of two functions is the sum of the derivatives of each function

$\frac{d}{dx}\left(xy^3\right)+\frac{d}{dx}\left(yx^3\right)=0$
4

The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function

$y^3\frac{d}{dx}\left(x\right)+y\frac{d}{dx}\left(x^3\right)=0$
5

The derivative of the linear function is equal to $1$

$1y^3+y\frac{d}{dx}\left(x^3\right)=0$
6

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$1y^3+3yx^{2}=0$
7

Any expression multiplied by $1$ is equal to itself

$y^3+3yx^{2}=0$

Answer

$y^3+3yx^{2}=0$

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Problem Analysis

Main topic:

Differential calculus

Time to solve it:

0.2 seconds

Views:

113