Use implicit differentiation to find dydx , if x3y5 + 3x2 = 8y3 + 1

Sure, let's use implicit differentiation to find dy/dx for the given equation x^3*y^5 + 3x^2 = 8y^3 + 1.

Step 1: Differentiate both sides of the equation with respect to x. Remember to apply the product rule (d/dx[uv] = u'v + uv') and the chain rule (d/dx[f(g(x))] = f'(g(x))g'(x)) where necessary.

The derivative of the left side of the equation is:

d/dx[x^3y^5 + 3x^2] = 3x^2y^5 + x^35y^4(dy/dx) + 6x.

The derivative of the right side of the equation is:

d/dx[8y^3 + 1] = 24y^2*(dy/dx).

Step 2: Set the derivatives equal to each other and solve for dy/dx.

3x^2y^5 + x^35y^4*(dy/dx) + 6x = 24y^2*(dy/dx).

Step 3: Move the terms involving dy/dx to one side of the equation and the remaining terms to the other side.

x^35y^4(dy/dx) - 24y^2*(dy/dx) = -3x^2*y^5 - 6x.

Step 4: Factor out dy/dx.

dy/dx * [5x^3y^4 - 24y^2] = -3x^2y^5 - 6x.

Step 5: Solve for dy/dx.

dy/dx = [-3x^2y^5 - 6x] / [5x^3y^4 - 24y^2].

So, the derivative of y with respect to x, dy/dx, for the given equation is [-3x^2y^5 - 6x] / [5x^3y^4 - 24y^2].