Question

Differentiate with respect to t
Assume x = x(t)
Assume y = y(t)
Assume z = z(t)

x^2 - y^3 + z^4 = 1

Answer 1

Answer:

$\displaystyle 2x\frac{dx}{dt}-3y^2\frac{dy}{dt}+4z^3\frac{dz}{dt}=0$

Step-by-step explanation:

We want to differentiate the equation:

$x^2-y^3+z^4=1$

With respect to t, where x, y, and z are functions of t.

So:

$\displaystyle \frac{d}{dt}\left[x^2-y^3+z^4\right]=\frac{d}{dt}\left[1\right]$

Implicitly differentiate on the left. On the right, the derivative of a constant is simply zero. Hence:

$\displaystyle 2x\frac{dx}{dt}-3y^2\frac{dy}{dt}+4z^3\frac{dz}{dt}=0$