find the ∂z/∂x and ∂z/∂y derivatives for the function: x^2sin(2y - 5z) = 1 + ycos(6zx)

Question

maybe implicit derivative? im not sure about that can u guys help me how to solve that problem?

Answer 1

Yes, it is about implicit derivative.

For ∂z/∂x part, you need to take derivative with respect to x at both sides.

The left-hand side is

∂(x²sin(2y-5z))=∂(x²)sin(2y-5z)+x²∂(sin(2y-5z)).

Here we used the product rule. Then, we have

LHS=2x∗sin(2y-5z)+x²cos(2y-5z)∗(-5∂z/∂x)

=2x∗sin(2y-5z)-5x²cos(2y-5z)∗(∂z/∂x).

Here we used the chain rule and ∂y/∂x=0, namely ∂(sin(2y-5z))=cos(2y-5z)∗∂(2y-5z)=cos(2y-5z)∗(-5∂z/∂x).

The right-hand side is similar. We have

∂(1+ycos(6zx))=∂(1)+∂(ycos(6zx))=∂(y)∗cos(6zx)+y∗∂(cos(6zx))

=y∗(-sin(6zx))∗∂(6zx)=y∗(-sin(6zx))∗6(∂(z)∗x+z∗∂(x))

=y∗(-sin(6zx))∗6((∂z/∂x)∗x+z)

=-6y∗sin(6zx)∗((∂z/∂x)∗x+z)

After the computation, we have

2x∗sin(2y-5z)-5x²cos(2y-5z)∗(∂z/∂x)=-6y∗sin(6zx)∗((∂z/∂x)∗x+z).

To find ∂z/∂x, we just need the simplify the above equation and solve for ∂z/∂x. We move the terms containing ∂z/∂x to the LHS and the rest to the RHS,

6y∗sin(6zx)∗(∂z/∂x)-5x²cos(2y-5z)∗(∂z/∂x)=-2x∗sin(2y-5z)-6y∗sin(6zx)∗z.

Factor out ∂z/∂x,

(6y∗sin(6zx)-5x²cos(2y-5z))*(∂z/∂x)=-2x∗sin(2y-5z)-6y∗sin(6zx)∗z,

and divide both sides by the factor in front of ∂z/∂x,

∂z/∂x = (-2x∗sin(2y-5z)-6y∗sin(6zx)∗z)/(6y∗sin(6zx)-5x²cos(2y-5z)).

You can try the same thing to find ∂z/∂y.