Question Video: Differentiating a Combination of Root Functions | Nagwa Question Video: Differentiating a Combination of Root Functions | Nagwa

# Question Video: Differentiating a Combination of Root Functions Mathematics • Second Year of Secondary School

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Find d/dπ₯(β2β(π₯) β 7π₯).

02:50

### Video Transcript

Find the derivative of negative two times the square root of π₯ minus seven π₯ with respect to π₯.

Weβre asked to find the derivative of an expression with respect to π₯. And we know how to differentiate negative seven π₯ with respect to π₯. However, we donβt know how to differentiate negative two root π₯ directly. So to differentiate this expression, weβre going to want to rewrite root π₯ in a form which we can differentiate. Luckily, we can do this by using our laws of exponents.

We recall the square root of π₯ is the same as π₯ to the power of one-half. So by using this, we can rewrite this as the derivative of negative two π₯ to the power of one-half minus seven π₯ with respect to π₯. And now we can see we can differentiate this term by term by using the power rule for differentiation. We recall this tells us for any real constants π and π, the derivative of ππ₯ to the πth power with respect to π₯ is equal to π times π times π₯ to the power of π minus one. We multiply by our exponent of π₯ and then reduce this exponent by one. And we want to apply this to each term separately.

In our first term, we want our value of π equal to negative two and our value of π equal to one-half. So by using the power rule for differentiation, we multiply by our exponent one-half and then reduce this exponent by one. This gives us negative two times one-half multiplied by π₯ to the power of one-half minus one.

We then want to apply this to our second term. And it might be easier to think of negative seven π₯ as negative seven π₯ to the first power. Then we can see that our value of π is negative seven and the value of our exponent of π₯ is one. So applying the power rule for differentiation on our second term, we want to multiply by our exponent of π₯, which is one, and then reduce this exponent by one. This gives us negative seven times one multiplied by π₯ to the power of one minus one.

Now, we can start simplifying this expression. First, the coefficient in our first term, negative two times one-half, is equal to negative one. And our exponent one-half minus one is equal to negative one-half. So our first term is negative π₯ to the power of negative one-half. Then, in our second term, the coefficient negative seven times one is equal to negative seven. And then in our exponent of π₯, we have one minus one, which is equal to zero. And of course, instead of adding negative seven times π₯ to the zeroth power, we can instead just subtract seven π₯ to the zeroth power. But then π₯ to the zeroth power is just equal to one. So we can just remove this altogether.

We could leave our answer like this. However, we can rewrite π₯ to the negative one-half by using our laws of exponents. We need to recall that raising a number to the negative exponent is the same as dividing by that same number raised to the positive exponent. So π₯ to the power of negative one-half is the same as dividing by π₯ to the power of one-half, which is of course equal to one divided by root π₯.

So by using our laws of exponents and rearranging our two terms, we got negative seven minus one divided by root π₯, which is our final answer. Therefore, by using our laws of exponents and the power rule for differentiation, we were able to show the derivative of negative two root π₯ minus seven π₯ with respect to π₯ is equal to negative seven minus one divided by root π₯.

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