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Katarina [22]
2 years ago
9

Differentiate with respect to t

Mathematics
1 answer:
True [87]2 years ago
8 0

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 <em>t</em>, where <em>x, y, </em>and <em>z</em> are functions of <em>t. </em>

<em />

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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