Find dy/dx of the given equation?

7x² + 6x^4 - 3y³ + sin(y) = 4

Differentiate Implicitly:

14x + 24x³ - 9y² y' + cos(y) y' = 0

(cos(y) - 9y²) y' = -2x(7 + 12x²)

y' = -2x(7 + 12x²) / (cos(y) - 9y²)

Alternatively, using partial derivatives: fx(x,y) and fy(x,y)

y' = -fx / fy

y' = -(14x + 24x³) / (cos(y) - 9y²)

y' = -2x(7 + 12x²) / (cos(y) - 9y²)

First two terms: use the power rule d(x^n)/dx = n * x^(n-1)

Third term: Power rule plus chain rule. It's 9y^2 * dy/dx. The 9y^2 comes from the power rule, the dy/dx is because of the chain rule.

Fourth term: Use the chain rule.

Right hand side: derivative of a constant is 0.

using implicit differentiation you take the derivative of both sides of the equation. i'm going to use y' instead of dy/dx but they both mean the same.

d/dx 7x^2+6x^4-3y^3+sin(y) = d/dx 4

14x+24x^3-9y^2y'+cosyy' = 0 (hope u got how i differentiated each term of the equation)

so u now solve this equation for y'(which is dy/dx)

24x^3+14x = 9y^2y'-cosyy'

24x^3+14x = y'(9y^2-cosy)

divide both sides by 9y^2-cosy

y' = 24x^3+14x/9y^2-cosy

dy/dx = (24x^3 + 14x) / (9y^2 - cosy)

we use implicit differentiation because we cannot solve for y

y=y(x) , then implicit d/dx

14x +24x^3 -9y^2 dy/dx +cosy dy/dx=0

dy/dx= (14x+24x^3 )/ (9y^2 -cos y)

At x=xo , you must find yo from the equation .- For solving you need to use a Numerical Method.-

Then plug (xo,yo) at dy/dx .-

7x^2+6x^4-3y^3+sin(y)=4

14x+24x^3-9y^2dy/dx+cos(y)dy/dx=0

-9y^2dy/dx+cos(y)dy/dx=-14x-24x^3

(dy/dx)(cos(y)-9y^2)=-14x-24x^3

dy/dx=(-14x-24x^3)/(cos(y)-9y^2)

dy/dx=2x(7+12x^2)/(9y^2-cos(y))