Video: Differentiating a Combination of Root Functions

Differentiate 𝑦 = √π‘₯ (5π‘₯√π‘₯ βˆ’ (4π‘₯/√π‘₯) βˆ’ 1).

03:49

Video Transcript

Differentiate 𝑦 equals root π‘₯ multiplied by five π‘₯ root π‘₯ minus four π‘₯ over root π‘₯ minus one.

So, the first thing we’re gonna do before we differentiate is rewrite our function in exponent form. And to do that, we’re gonna use a couple of exponent laws. So, we’ve got π‘₯ to the power of a half is equal to root π‘₯. π‘₯ to the power of negative π‘Ž is equal to one over π‘₯ to the power of π‘Ž. And when we do that, we’re gonna get 𝑦 is equal to π‘₯ to the power of a half multiplied by five π‘₯ multiplied by π‘₯ to the power of a half minus four π‘₯ multiplied by π‘₯ to the power of negative a half minus one.

So, now to simplify it further, what we’re going to do is we’re gonna use another exponent law. And that one is that if we have π‘₯ to the power of π‘Ž multiplied by π‘₯ to the power of 𝑏, then it’s equal to π‘₯ to the power of π‘Ž plus 𝑏. So, we add the exponents.

So, then we got 𝑦 is equal to π‘₯ to the power of a half multiplied by five π‘₯ to the power of three over two. That’s because we added the exponents. So, we had one add a half, so it gives us one and a half, or three over two, minus four π‘₯ to the power of a half minus one. And we got the four π‘₯ to the power of a half because we had four π‘₯ multiplied by π‘₯ to the power of negative a half. So, that’s adding the exponents. So, one add negative a half, which is the same as one take away a half, which is just a half.

Now the final stage for simplifying before we differentiate is to expand the parentheses. So, we have π‘₯ to the power of a half multiplied by five π‘₯ to the power of three over two. So, when we multiply these together, we’re gonna get five π‘₯ squared. That’s because if we add the exponents, we’ll have a half plus three over two, or three-halves, which gives us four-halves, which is the same as two.

And then the second term is going to be four π‘₯. That’s cause we have π‘₯ to the power of a half multiplied by four π‘₯ to the power of a half. So, again, we add the exponents. A half plus a half is one. And then, finally, we have π‘₯ to the power of a half. So, great! We now have it in a form that’s gonna be easy to differentiate.

So, we can now say that the derivative of our function, so 𝑦 prime, or can be written as 𝑑𝑦 𝑑π‘₯, it’s gonna be equal to 10π‘₯ for our first term. That’s cause we have the exponent multiplied by the coefficients, so two multiplied by five, which is 10. And then we reduce the exponent by one. So, we go from two to one, so we get 10π‘₯. And then, we get minus four. That’s because, in the same way, we differentiate four π‘₯. So, we multiply the exponent which is one by four, which gives us four, then reduce the exponent to zero. So, we’re just left with four. Because π‘₯ to the power of zero is just one.

And finally, the last term is negative a half π‘₯ to the power of negative a half. And again, that’s because we had a half multiplied by one, which gives us a half. Then we reduce the exponent by one, so we go from a half to negative a half.

So, now finally, we’re just gonna rewrite our answer back in its original form. And we’re gonna do that by using our exponent laws to rewrite the final term. So therefore, we can say that the derivative of 𝑦 equals root π‘₯ multiplied by five π‘₯ root π‘₯ minus four π‘₯ over root π‘₯ minus one is going to be 10π‘₯ minus four minus one over two root π‘₯. And we got that because we had π‘₯ to the power of negative a half. Well, we know that if we have π‘₯ to the power of a half, it’s gonna give us root π‘₯. But then we use the law that tells us that if we have π‘₯ to the power of negative π‘Ž, it’s one over π‘₯ to the power π‘Ž. So, that’s how we write our final answer.

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