# Question Video: Finding the First Derivative of Functions Involving Trigonometric Ratios and Roots Using the Chain Rule Mathematics • Higher Education

Find the first derivative of the function 𝑦 = √(8𝑥 − sin⁸ 9𝑥).

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### Video Transcript

Find the first derivative of the function 𝑦 equals the square root of eight 𝑥 minus sin of nine 𝑥 to the power of eight.

Here we have 𝑦 is equal to the square root of another function, so we have a composite function. We’re, therefore, going to apply the chain rule. We’re going to define 𝑢 to be the function underneath the square root, so 𝑢 is equal to eight 𝑥 minus sin of nine 𝑥 to the power of eight. Then, 𝑦 is equal to the square root of 𝑢, which we can express using index notation as 𝑢 to the power of one-half.

The chain rule tells us that d𝑦 by d𝑥 is equal to d𝑦 by d𝑢 multiplied by d𝑢 by d𝑥. So, we need to find each of these derivatives. d𝑦 by d𝑢 is relatively straightforward. Using the power rule, we get one-half 𝑢 to the power of negative one-half. For d𝑢 by d𝑥, the derivative of eight 𝑥 is just eight. But what about the derivative of sin of nine 𝑥 to the power of eight?

We actually need to apply the chain rule again. We can let 𝑔 equal this function. And we can change the notation a little to write it as sin nine 𝑥 to the power of eight. It’s an equivalent notation, but it might it make it a little clearer how we’re going to find the derivative.

We recall the chain rule extension to the power rule, which told us that if we had a function 𝑓 of 𝑥 raised to a power 𝑛, then its derivative was 𝑓 prime of 𝑥 multiplied by 𝑛 multiplied by 𝑓 of 𝑥 to the power of 𝑛 minus one. Here, we have a function, sin of nine 𝑥 raised to a power eight, so we can apply the chain rule extension to the power rule. We need to recall one more rule which is that the derivative with respect to 𝑥 of sin 𝑎𝑥 is 𝑎 cos 𝑎𝑥.

So, we begin. The derivative of the part inside the parentheses is nine cos nine 𝑥. Then, we multiply by the power eight. And then, we have the function inside the parentheses written out again, but with the power reduced by one. Simplifying gives 72 cos nine 𝑥 sin to the power of seven nine 𝑥. So, now that we found both d𝑦 by d𝑢 and d𝑢 by d𝑥, we can substitute into the chain rule.

We have then that d𝑦 by d𝑥 is equal to a half 𝑢 to power of negative a half multiplied by eight minus 72 cos nine 𝑥 sin nine 𝑥 to the power of seven. We must also remember to replace 𝑢 in terms of 𝑥. So, 𝑢 is equal to eight 𝑥 minus sin of nine 𝑥 to the power of eight. We’ll also simplify the fractions. Dividing by that denominator of two leaves coefficients of four and 36 in the numerator.

And we recall also that 𝑢 to the power of negative a half is equal to one over root 𝑢. So, our derivative d𝑦 by d𝑥 simplifies to four minus 36 cos nine 𝑥 sin nine 𝑥 to the power of seven all over the square root of eight 𝑥 minus sin nine 𝑥 to the power of eight.

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