# 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 π¦ = β(8π₯ β sinβΈ 9π₯).

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### Video Transcript

Find the first derivative of the function π¦ equals the square root of eight π₯ minus sin of nine π₯ to the power of eight.

Here we have π¦ is equal to the square root of another function, so we have a composite function. Weβre, therefore, going to apply the chain rule. Weβre going to define π’ to be the function underneath the square root, so π’ is equal to eight π₯ minus sin of nine π₯ to the power of eight. Then, π¦ is equal to the square root of π’, which we can express using index notation as π’ to the power of one-half.

The chain rule tells us that dπ¦ by dπ₯ is equal to dπ¦ by dπ’ multiplied by dπ’ by dπ₯. So, we need to find each of these derivatives. dπ¦ by dπ’ is relatively straightforward. Using the power rule, we get one-half π’ to the power of negative one-half. For dπ’ by dπ₯, the derivative of eight π₯ is just eight. But what about the derivative of sin of nine π₯ to the power of eight?

We actually need to apply the chain rule again. We can let π equal this function. And we can change the notation a little to write it as sin nine π₯ to the power of eight. Itβs an equivalent notation, but it might it make it a little clearer how weβ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 π of π₯ raised to a power π, then its derivative was π prime of π₯ multiplied by π multiplied by π of π₯ to the power of π minus one. Here, we have a function, sin of nine π₯ 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 π₯ of sin ππ₯ is π cos ππ₯.

So, we begin. The derivative of the part inside the parentheses is nine cos nine π₯. 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 π₯ sin to the power of seven nine π₯. So, now that we found both dπ¦ by dπ’ and dπ’ by dπ₯, we can substitute into the chain rule.

We have then that dπ¦ by dπ₯ is equal to a half π’ to power of negative a half multiplied by eight minus 72 cos nine π₯ sin nine π₯ to the power of seven. We must also remember to replace π’ in terms of π₯. So, π’ is equal to eight π₯ minus sin of nine π₯ to the power of eight. Weβ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 π’ to the power of negative a half is equal to one over root π’. So, our derivative dπ¦ by dπ₯ simplifies to four minus 36 cos nine π₯ sin nine π₯ to the power of seven all over the square root of eight π₯ minus sin nine π₯ to the power of eight.