# Question Video: Differentiating a Combination of Root Functions Mathematics • Higher Education

Evaluate (𝑑/d𝑥) (−5/∛𝑥).

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### Video Transcript

Evaluate 𝑑 𝑑𝑥 of negative five over the cube root of 𝑥.

So in this question, we’re asked to differentiate negative five over the cube root of 𝑥. But the first thing we’re gonna do is rewrite it in exponent form. And to help us rewrite it, we’ve got a couple of exponent laws that we’re gonna use. The first is that if we have 𝑥 to the power of a third, then this is equal to the cube root of 𝑥. And we also have 𝑥 to the power of negative 𝑎 is equal to one over 𝑥 to the power of 𝑎.

So therefore, using these laws, what we’re gonna get is negative five 𝑥 to the power of negative a third. And that’s because we had the cube root of 𝑥, which we already said was equal to 𝑥 to the power of a third. However, because it was five over the cube root of 𝑥, then it makes our exponent negative. So it will be 𝑥 to the negative a third.

So now, if we’re gonna differentiate this, what we’re gonna do is, first of all, we’ll multiply the exponent by the coefficient. So we have negative a third multiplied by negative five. And then, we’ve got 𝑥 to the power of — we’ve got negative a third minus one because we subtracted one from the exponent. It’s worth remembering we’re about to do this, that one would be the same as three over three or three-thirds. So this is gonna give us five-thirds 𝑥 to the power of negative four-thirds. That’s because we had negative a third multiplied by negative five. Well, a negative multiplied by a negative is a positive. And then, we had negative a third minus three-thirds. So it gives us negative four-thirds.

So now what we’re gonna do is rewrite in the original form using the same rules that we used before but in reverse. And this is gonna be equal to five over three multiplied by the cube root of 𝑥 to the power of four. And this was from the combination of the rules we had earlier. Or we can think if we’re using another law, that’s if we had 𝑥 to the power of 𝑎 over 𝑏, then this is equal to the 𝑏th root of 𝑥 to the power of 𝑎. Or we can write it as the 𝑏th root of 𝑥 all to the power of 𝑎.