# Question Video: Differentiating Functions Involving Reciprocal Trigonometric Ratios and Roots Using the Chain Rule Mathematics • Higher Education

If π¦ = β(19 csc π₯ + 18), find dπ¦/dπ₯.

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### Video Transcript

If π¦ is equal to the square root of 19 csc of π₯ plus 18, find dπ¦ by dπ₯.

Weβre given π¦ as a function in π₯, and weβre asked to find dπ¦ by dπ₯. Thatβs the first derivative of π¦ with respect to π₯. So we need to differentiate our expression for π¦ with respect to π₯. Looking at this, we can see this is the square root of a trigonometric function. In other words, π¦ is the composition of functions. And to differentiate the composition of functions, we can do this by using the chain rule, and this would work. However, because our outer function is a power function, we can actually do this by using the general power rule. It doesnβt matter which method we would use; both would give us the same answer; itβs personal preference. In this case, weβre going to use the general power rule.

To use the general power rule, weβre first going to need to use our laws of exponents to rewrite π¦ as 19 csc of π₯ plus 18 all raised to the power of one-half. Letβs now recall the general power rule. This tells us for any real constant π and differentiable function π of π₯, if π¦ is equal to π of π₯ all raised to the πth power, then dπ¦ by dπ₯ would be equal to π times π prime of π₯ multiplied by π of π₯ all raised to the power of π minus one. And we can see that π¦ is exactly written in this form, with our exponent π equal to one-half and π of π₯, our inner trigonometric function, 19 csc of π₯ plus 18.

However, to use the general power rule, we need to first find an expression for π prime of π₯. And π prime of π₯ will be the derivative of 19 csc of π₯ plus 18 with respect to π₯. And thereβs a few different ways we could evaluate this derivative. For example, we could rewrite the csc of π₯ by using our Pythagorean identity to be one over the sin of π₯. We could then differentiate this by using the general power rule or the chain rule, or we could do this by using the quotient rule. And any of those methods would work. However, we can also recall the following result for differentiating reciprocal trigonometric functions.

The derivative of the csc of π₯ with respect to π₯ is equal to negative the cot of π₯ multiplied by the csc of π₯. We can then use this to differentiate π of π₯ term by term. First, the derivative of 19 csc of π₯ with respect to π₯ will be negative 19 cot of π₯ times the csc of π₯. Next, we know the derivative of the constant 18 with respect to π₯ will just be equal to zero.

Now that weβve found an expression for π prime and we know the value of π and an expression for π of π₯, we can use the general power rule to find dπ¦ by dπ₯. Substituting in our expressions for π of π₯, π prime of π₯, and π is equal to one-half into our general power rule, we get dπ¦ by dπ₯ is equal to one-half times negative 19 cot of π₯ times the csc of π₯ multiplied by 19 csc of π₯ plus 18 all raised to the power of one-half minus one.

And we can simplify this. First, in our exponent, one-half minus one is equal to negative one-half. We could then simplify our coefficient. However, recall from our laws of exponents, π to the power of negative one-half is equal to one divided by the square root of π. So instead of multiplying by 19 csc of π₯ plus 18 all raised to the power of negative one-half, we can instead divide by the square root of 19 csc of π₯ plus 18.

So by doing this and rearranging slightly, we get our final answer. Therefore, by using the general power rule, we were able to show if π¦ is equal to the square root of 19 csc of π₯ plus 18, then dπ¦ by dπ₯ will be equal to negative 19 cot of π₯ times the csc of π₯ all divided by two times the square root of 19 csc of π₯ plus 18.