Question Video: Differentiating Root Functions

Evaluate d/dπ‘₯ (βˆ’5π‘₯^(1/9)).

02:20

Video Transcript

Evaluate the derivative of negative five times π‘₯ to the power of one over nine with respect to π‘₯.

In this question, we’re asked to evaluate the derivative of an expression, and we can see that this is in a special form. It’s in the form π‘Ž times π‘₯ to the 𝑛th power where π‘Ž and 𝑛 are real constants. And we know we can evaluate derivatives of this form by using the power rule for differentiation. The power rule for differentiation tells us 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.

So to differentiate negative five times π‘₯ to the power of one over nine, we’ll set our constant coefficient π‘Ž equal to negative five and our exponent of π‘₯, 𝑛, equal to one over nine. We can then substitute these values into our power rule for differentiation to evaluate its derivative. So, substituting π‘Ž is equal to negative five and 𝑛 is equal to one over nine into our power rule for differentiation, we can evaluate this derivative to give us negative five times one over nine multiplied by π‘₯ to the power of one ever nine minus one. And of course, we can simplify this. First, our coefficient negative five times one over nine is equal to negative five over nine. Next, in our exponent of π‘₯, we have one-ninth minus one is equal to negative eight over nine.

So, we were able to rewrite the derivative of negative five π‘₯ to the power of one over nine with respect to π‘₯ as negative five over nine times π‘₯ to the power of negative eight over nine. And we could just leave our answer like this. However, we can also rewrite this by using our laws of exponents. We need to recall that π‘₯ to the power of negative π‘š divided by 𝑛 will be equal to one divided by the 𝑛th root of π‘₯ to the power of π‘š.

This is because raising a number to a negative exponent is the same as dividing by that number raised to the positive exponent. Then, in our denominator, we have π‘₯ to the power of π‘š over 𝑛 will be equal to π‘₯ to the power of 𝑛 all raised to the power of one over 𝑛. And then, we know that raising a number to the power of one over 𝑛 is the same as finding the 𝑛th root of that number. And so, by using our laws of exponents, we can rewrite this as negative five divided by nine times the ninth root of π‘₯ to the eighth power, which is our final answer.

Therefore, by using the power rule for differentiation and our laws of exponents, we were able to show the derivative of negative five times π‘₯ to the power of one over nine with respect to π‘₯ is equal to negative five divided by nine times the ninth root of π‘₯ to the eighth power.

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