Video: Differentiating Root Functions

Find d𝑦/dπ‘₯, given that 𝑦 = βˆ›(π‘₯).

02:44

Video Transcript

Find d𝑦 by dπ‘₯, given that 𝑦 is equal to the cube root of π‘₯.

We’re given that 𝑦 is the cube root of π‘₯, and we’re asked to find d𝑦 by dπ‘₯. That’s the derivative of 𝑦 with respect to π‘₯. So we need to differentiate this expression for 𝑦 with respect to π‘₯. However, we don’t know how to do this in its current form. So we’re going to want to rewrite this into an expression which we do know how to differentiate. And luckily, we can do this by using our laws of exponents. We know for any real constant π‘Ž, the cube root of π‘Ž will be equal to π‘Ž to the power of one over three. This means we can rewrite the cube root of π‘₯ as π‘₯ to the power of one over three. And we can then differentiate π‘₯ to the power of one over three by using the power rule for differentiation.

So because we showed that 𝑦 is equal to π‘₯ to the power of one-third, d𝑦 by dπ‘₯ will be equal to the derivative of π‘₯ to the power of one over three with respect to π‘₯. And we can evaluate the derivative of this by using the power rule for differentiation. We recall this tells us for any real constants π‘Ž and 𝑛, the derivative of π‘Žπ‘₯ to the 𝑛th power with respect to π‘₯ is equal to π‘Ž times 𝑛 times π‘₯ to the power of 𝑛 minus one. We multiply by our exponent of π‘₯ and then reduce this exponent by one.

In our case, we can see the constant coefficient of π‘₯ is equal to one and our exponent of π‘₯ is equal to one-third. So we’ll set our value of π‘Ž equal to one and 𝑛 equal to one-third. So when we multiply by our exponent of π‘₯ and reduce this exponent by one, we’ll get one-third multiplied by π‘₯ to the power of one over three minus one. And of course we can simplify this. In our exponent, one-third minus one is equal to negative two over three. So we were able to show that d𝑦 by dπ‘₯ is equal to one-third times π‘₯ to the power of negative two over three. And we could leave our answer like this. However, we can also simplify this by using our laws of exponents.

First, we recall that π‘Ž to the power of negative π‘š over 𝑛 will be the same as one divided by π‘Ž to the power of π‘š over 𝑛. So by setting π‘Ž equal to π‘₯, π‘š equal to two, and 𝑛 equal to three, we can rewrite our expression as one-third times one over π‘₯ to the power of two over three. However, we can simplify this even further. Recall, π‘Ž to the power of π‘š over 𝑛 will be equal to π‘Ž to the power of π‘š all raised to the power of one over 𝑛. But we can then combine this with another one of our laws of exponents. Raising a number to the power of one over 𝑛 is the same as taking the 𝑛th root of that number.

So we can replace π‘₯ to the power of two over three with the cube root of π‘₯ squared. This gives us the following expression. And there’s one last piece of manipulation we’ll do. We need to notice the cube root of one is equal to one. This means that one divided by the cube root of π‘₯ squared is equal to the cube root of one over π‘₯ squared. So we’ll replace this in our expression, and this gives us our final answer. Therefore, by using the power rule for differentiation and our rules for exponents, we were able to show if 𝑦 is equal to the cube root of π‘₯, then d𝑦 by dπ‘₯ will be equal to one-third times the cube root of one over π‘₯ squared.

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