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Can someone clarify what it means to find the derivative "with respect to $t$"?

For example $$x^2 + y^2 = 625.$$

Using implicit differentiation I get: $2x + 2y(dy/dx) = 0$.

However with differentiation "respect to $t$" the answer is: $2x(dx/dt) + 2y(dy/dt) = 0$

Does it just mean multiplying it with $d/dt$ after differentiating it normally?

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1 Answer

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It means you have to think of $x$ implicitly as a function of $t$ rather than an independent variable. Same thing with $y$. So we can rewrite your original equation as $x(t)^2 + y(t)^2 = 625$, then just differentiate both sides of the equation with respect to $t$. The right hand side will become $0$ after differentiating. The derivative of $x(t)^2$ is, by the chain rule, $2x(t)\cdot (dx/dt)$ and similarly for $y(t)^2$, so the equation becomes $2x(t)\cdot (dx/dt) + 2y(t)\cdot (dy/dt) = 0$.

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