Differentiate with respect to t Assume x = x(t) Assume y = y(t) Assume z = z(t) x^2 – y^3 + z^4 = 1

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

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niczorrrr 4 years 2021-08-12T06:52:26+00:00 1 Answers 11 views 0

Answers ( )

  1. huyenthanh
    0
    2021-08-12T06:53:40+00:00
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    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

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