Question

a) Find dy/dx if x5 +xy3 +x2y+y4=4
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Answer 1

Hope the pic i attached below helps you!!

Answer 2

Concept:-

It might resemble a word or a number representation of the quantity's arithmetic value. It could resemble a word or a number that represents the numerical value of the quantity. It could have the appearance of a word or a number that denotes the quantity's numerical value.

Given:-

We have been given the expression that "$x^5+ xy^3 + x^2y + y^4 =4$"

Find:-

We need to find the value of $dy/dx$ from the given equation "$x^5+ xy^3 + x^2y + y^4 =4$".

Solution:-

The given equation in the question: $x^5+ xy^3 + x^2y + y^4 =4$

Differentiating with respect to $dx$ on both sides of the equation, we get the following result

$\Rightarrow d( x^5 + xy^3 + x^2y + y^4) / dx = d ( 4 ) / dx\\\Rightarrow d ( x^5 ) / dx + d ( xy^3 ) / dx + d ( x^2y ) + d ( y^4 ) / dx = d ( 4 ) / dx\\\Rightarrow 5x^4 + { x. d ( y^3)/dx + y^3 dx/dx } + { x^2 dy/dx + y d ( x^2)/dx } + d ( y^4) / dx = 0\\\Rightarrow 5x^4 + x. ( 3y^2) dy/dx + y^3 + x^2 dy/dx + y. ( 2x ) + 4y^3 dy/dx = 0\\\Rightarrow 5x^4 + 3xy^2 dy/dx + y^3 + x^2 dy/dx + 2xy +4y^3 dy/dx = 0$

$\Rightarrow 5x^4 + y^3+ ( 3xy^2+ x^2+ 4y^3) dy/dx = 0\\\Rightarrow ( 3xy^2 + x^2 + 4y^3) dy/dx = - ( 5x^4+ y^3)\\\Rightarrow dy/dx = - ( 5x^4+ y^3) / ( 3xy^2+ x^2+ 4y^3)$

Hence, the value of $dy/dx$ is $- ( 5x^4+ y^3 ) / ( 3xy^2+ x^2 + 4y^3 )$.

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