# Question Video: Instantaneous Rates of Change Mathematics • Higher Education

Find the instantaneous rate of change of π(π₯) = βπ₯ at π₯ = π₯β > 0.

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### Video Transcript

Find the instantaneous rate of change of π of π₯ is equal to the square root of π₯ at π₯ equals π₯ one, which is greater than zero.

Remember, the instantaneous rate of change of a function π of π₯ at a point π₯ equals π is found by taking the limit as β approaches zero of the average rate of change function. Thatβs the limit as β approaches zero of π of π plus β minus π of π all over β. In this case, we know that π of π₯ is equal to the square root of π₯, and we want to find the instantaneous rate of change at π₯ equals π₯ one. So, weβll let π be equal to π₯ one. Letβs substitute what we know into our formula. We want to compute the limit as β approaches zero of π of π₯ one plus β minus π of π₯ one all over β. We need to find the limit as β approaches zero of the square root of π₯ one plus β minus the square root of π₯ one all over β.

Now, we canβt do this with direct substitution. If we do, we end up dividing by zero and we know that to be undefined. And so instead, we multiply the numerator and denominator of the function by the conjugate of the numerator, by the square root of π₯ plus one plus β plus the square root of π₯ one. On the denominator, we simply have β times the square root of π₯ one plus β plus the square root of π₯ one. Then on the numerator, we have the square root of π₯ one plus β times the square root of π₯ one plus β, which is simply π₯ one plus β. Then, we multiply the square root of π₯ one plus β by the square root of π₯, and negative the square root of π₯ one times the square root of π₯ one plus β. When we find their sum, we get zero.

So, all thatβs left to do is to multiply negative the square root of π₯ one by the square root of π₯ one. And we simply get negative π₯ one. π₯ one minus π₯ one is zero. And then, we divide through by β. And so, this becomes the limit as β approaches zero of one over the square root of π₯ one plus β plus the square root of π₯ one. And we can now evaluate this as β approaches zero. Weβre left with one over the square root of π₯ one plus the square root of π₯ one, which is one over two times the square root of π₯ one. The instantaneous rate of change function of π of π₯ is equal to the square root of π₯ is therefore one over two times the square root of π₯ one.