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 = √(8π‘₯ βˆ’ sin⁸ 9π‘₯).

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Video Transcript

Find the first derivative of the function 𝑦 equals the square root of eight π‘₯ minus sin of nine π‘₯ 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 𝑒 to be the function underneath the square root, so 𝑒 is equal to eight π‘₯ minus sin of nine π‘₯ to the power of eight. Then, 𝑦 is equal to the square root of 𝑒, which we can express using index notation as 𝑒 to the power of one-half.

The chain rule tells us that d𝑦 by dπ‘₯ is equal to d𝑦 by d𝑒 multiplied by d𝑒 by dπ‘₯. So, we need to find each of these derivatives. d𝑦 by d𝑒 is relatively straightforward. Using the power rule, we get one-half 𝑒 to the power of negative one-half. For d𝑒 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 𝑔 equal this function. And we can change the notation a little to write it as sin nine π‘₯ 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 π‘₯ raised to a power 𝑛, then its derivative was 𝑓 prime of π‘₯ multiplied by 𝑛 multiplied by 𝑓 of π‘₯ to the power of 𝑛 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‘Žπ‘₯ is π‘Ž cos π‘Žπ‘₯.

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 d𝑦 by d𝑒 and d𝑒 by dπ‘₯, we can substitute into the chain rule.

We have then that d𝑦 by dπ‘₯ is equal to a half 𝑒 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 𝑒 in terms of π‘₯. So, 𝑒 is equal to eight π‘₯ minus sin of nine π‘₯ 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 𝑒 to the power of negative a half is equal to one over root 𝑒. So, our derivative d𝑦 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.

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