Use implicit differentiation to find 𝒅𝒚𝒅𝒙 if a) 5y2+siny= lnx2 b) 5y3-tany=ln2x-2x2

Sure, let's solve these problems one by one.

a) Given 5y^2 + sin(y) = ln(x^2)

First, differentiate both sides with respect to x. Remember that y is a function of x, so we need to use the chain rule when differentiating y.

The derivative of 5y^2 is 10y(dy/dx) and the derivative of sin(y) is cos(y)(dy/dx). On the right side, the derivative of ln(x^2) is 2/x.

So, we have 10y(dy/dx) + cos(y)(dy/dx) = 2/x.

Factor out dy/dx on the left side to get dy/dx(10y + cos(y)) = 2/x.

Finally, solve for dy/dx to get dy/dx = 2/(x(10y + cos(y))).

b) Given 5y^3 - tan(y) = ln(2x) - 2x^2

Again, differentiate both sides with respect to x. The derivative of 5y^3 is 15y^2(dy/dx), the derivative of tan(y) is sec^2(y)(dy/dx), the derivative of ln(2x) is 2/x, and the derivative of -2x^2 is -4x.

So, we have 15y^2(dy/dx) - sec^2(y)(dy/dx) = 2/x - 4x.

Factor out dy/dx on the left side to get dy/dx(15y^2 - sec^2(y)) = 2/x - 4x.

Finally, solve for dy/dx to get dy/dx = (2/x - 4x) / (15y^2 - sec^2(y)).