Question Video: Differentiating Root Functions Using the Chain Rule | Nagwa Question Video: Differentiating Root Functions Using the Chain Rule | Nagwa

# Question Video: Differentiating Root Functions Using the Chain Rule Mathematics • Second Year of Secondary School

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Find dπ¦/dπ₯, given that π¦ = β(13π₯)βΉ.

04:28

### Video Transcript

Find the derivative of π¦ with respect to π₯, given that π¦ is equal to the square root of 13π₯ all raised to the ninth power.

We need to find the derivative of π¦ with respect to π₯. And we can see that our function π¦ is in fact the composition of three functions. We take the linear function 13π₯, we raise it to the ninth power, and then we take the square root of this entire expression.

Since π¦ is written as the composition of functions which we know how to differentiate, we could do this by using the chain rule. However, currently, π¦ is the composition of three functions. So we would need to use the chain rule twice. However, we can simplify this expression. If we were to distribute nine over our parentheses, then we would have π¦ is the composition of two functions which we know how to differentiate.

In fact, we could then simplify even further by using our laws of exponents on the square root symbol. So letβs try simplifying our expression first before using the chain rule. In fact, it is worth pointing out there are two ways of simplifying this expression. We could use our laws of exponents to simplify the square root symbol first and then distribute our exponent over our parentheses.

However, weβll do this in the other order. Weβll start by distributing our exponent of nine over our parentheses. Weβll do this by using our laws of exponents. π times π all raised to the power of π is equal to π to the power of π multiplied by π to the power of π. So using this, we get π¦ is equal to the square root of 13 to the ninth power multiplied by π₯ to the ninth power. But we can simplify this even further.

We need to remember that the square root of π is equal to π to the power of one-half. So instead of taking the square root of 13 to the ninth power multiplied by π₯ to the ninth power, we can instead raise this entire expression to the power of one-half. But then we can just distribute our exponent over our parentheses once again. Each of the factors inside of our parentheses will now be raised to the power of one-half. This gives us 13 to the ninth power all raised to the power of one-half multiplied by π₯ to the ninth power all raised to the power of one-half.

And finally, we can use one more of our laws of exponents to simplify this expression. We remember that π to the power of π all raised to the power of π is equal to π to the power of π times π. So we have 13 to the ninth power all raised to the power of one-half is equal to 13 raised to the power of nine times one-half. And nine times one-half is equal to nine over two. Similarly, π₯ to the ninth power all raised to the power of one-half is equal to π₯ raised to the power of nine times one-half, which is π₯ to the power of nine over two.

So weβve now shown that π¦ is equal to 13 to the power of nine over two times π₯ to the power of nine over two. And now we can see why this is useful. 13 raised to the power of nine over two is a constant. And we can differentiate π₯ raised to the power of nine over two by using the power rule for differentiation.

So weβre now ready to find an expression for dπ¦ by dπ₯. Thatβs the derivative of 13 to the power of nine over two times π₯ to the power of nine over two with respect to π₯. And we can do this by using the power rule for differentiation. We want to multiply by our exponent of π₯ and then reduce this exponent by one. Our exponent of π₯ is nine over two. So we multiply by nine over two and then reduce the exponent by one. This gives us nine over two times 13 raised to the power of nine over two multiplied by π₯ to the power of nine over two minus one. And we can simplify this. First, in our exponent of π₯, nine over two minus one is just equal to seven over two.

Now, we could leave our answer like this. However, when we were given π¦, it was given as a function of 13π₯. So we could rewrite our answer in terms of 13π₯. So we want to try and write this in terms of 13π₯. To do this, we can notice that π₯ is raised to the power of seven over two. So we should write 13 to the power of seven over two, and then we can simplify this by using our laws of exponents.

So first, weβll start by splitting 13 to the power of nine over two into 13 multiplied by 13 to the power of seven over two. Now, we can see 13 is raised to the power of seven over two and π₯ is raised to the power of seven over two. So we can use our laws for exponents to rewrite this as 13π₯ all raised to the power of seven over two. This gives us nine over two times 13 multiplied by 13π₯ all raised to the power of seven over two. And the last thing weβll do is simplify nine over two multiplied by 13 to give us 117 divided by two. And this gives us our final answer.

Therefore, we showed if π¦ is equal to the square root of 13π₯ all raised to the ninth power, then dπ¦ by dπ₯ is equal to 117 over two times 13π₯ raised to the power of seven over two.

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