Use the chain rule to find dz/dt. z = xy³ - x²y, x = t² + 1, y = t² - 1

Solution:

The chain rule is used to find the derivatives of composite functions like (x² + 1)³, (sin 2x), (ln 5x), e^2x, and so on.

If y = f(g(x)), then y' = f'(g(x)). g'(x).

The chain rule states that the instantaneous rate of change of f relative to g relative to x helps us calculate the instantaneous rate of change of f relative to x

z = xy³ - x²y (1)

x = t² + 1; dx/dt = 2t

y = t² - 1 dy/dt = 2t

Differentiating equation (1) w.r.t 

dz/dt = y³dx/dt + 3y²x(dy/dt) - 2xydx/dt - x² dy/dt

= y³(2t) + 3y²x(2t) - 2xy(2t) - x²(2t)

= 2t[y³ + 3y²x - 2xy - x²]

= 2t[( t² - 1)³ + 3( t² - 1)²( t² + 1) - 2( t² + 1)(t² - 1) - ( t² + 1)²]

= 2t[t⁶ - 3t⁴ + 3t² - 1 + 3(t⁴ + 1 -2t²)( t² + 1) - 2(t⁴ - 1) - t⁴ - 2t² - 1]

= 2t[t⁶ - 3t⁴ + 3t² - 1 + 3(t⁶ + t⁴ + t² + 1 - 2 t⁴ - 2t²) - 2t⁴ + 2 - t⁴ - 2t² - 1]

= 2t[t⁶ - 3t⁴ + 3t² - 1 + 3t⁶ + 3t⁴ + 3t² + 3 - 6t⁴ - 6t² - 2t⁴ + 2 - t⁴ - 2t² - 1]

= 2t[4t⁶ - 9t⁴ - 2t² +3]

= 8t⁷ - 18t⁵ - 4t³ + 6t

---

Use the chain rule to find dz/dt.
z = xy³ - x²y, x = t² + 1, y = t² - 1

Summary:

Using the chain rule the value of dz/dt is 8t⁷ - 18t⁵ - 4t³ + 6t