Video: Differentiating the Composition of Trigonometric and Root Functions Using the Chain Rule

If 𝑦 = √(βˆ’4𝑧 + 9) and 𝑧 = sec 6π‘₯, find d𝑦/dπ‘₯ at π‘₯ = πœ‹/18.

05:50

Video Transcript

If 𝑦 is equal to the square root of negative four 𝑧 plus nine and 𝑧 is equal to the sec of six π‘₯, find the d𝑦 by dπ‘₯ at π‘₯ is equal to πœ‹ by 18.

The question wants us to find d𝑦 by dπ‘₯ which is the first derivative of 𝑦 with respect to π‘₯ when π‘₯ is equal to πœ‹ by 18. To do this, we need to first find an expression for d𝑦 by dπ‘₯ and then substitute π‘₯ is equal to πœ‹ by 18 into this expression. Normally, we would just do this by differentiating our expression for 𝑦 with respect to π‘₯. However, we can see a problem. We’re not given 𝑦 as a function in π‘₯. Instead, we’re given 𝑦 as a function in 𝑧.

And we can then see 𝑧 in turn is a function in π‘₯, so we can do this by using the chain rule. So, let’s start by recalling the chain rule. The chain rule tells us if 𝑦 is some function of 𝑧 and 𝑧 in turn is some function in π‘₯, then d𝑦 by dπ‘₯ will be equal to d𝑦 by d𝑧 multiplied by d𝑧 by dπ‘₯. And this is exactly what we have here. So, to find our expression for d𝑦 by dπ‘₯, we need to differentiate 𝑦 with respect to 𝑧 and 𝑧 with respect to π‘₯.

Let’s start by finding an expression for d𝑦 by d𝑧. That’s the derivative of the square root of negative four 𝑧 plus nine with respect to 𝑧. There’s a few different ways of differentiating this. For example, this is the composition of two functions. So, we could differentiate this by using the chain rule. However, we’re going to use the general power rule.

To do this, we first need to recall that taking the square root is the same as raising a number to the power for one-half. So, we can rewrite this as negative four 𝑧 plus nine all raised to the power of one-half. Then, we can evaluate this derivative by using the general power rule. We recall this tells us for the differentiable function 𝑔 of 𝑧 in any real constant 𝑛, the derivative of 𝑔 of 𝑧 all raised to the 𝑛th power with respect to 𝑧 is equal to 𝑛 times 𝑔 prime of 𝑧 multiplied by 𝑔 of 𝑧 all raised to the power of 𝑛 minus one.

We can do this directly. First, we need to multiply by our exponent 𝑛. In this case, we can see this is one-half. Next, in the power rule for differentiation, we need to differentiate our inner function. In this case, our inner function is a linear function. We know the derivative of this will be the coefficient of 𝑧, which is negative four. Finally, in the general power rule, we need to multiply it by our original function. However, we need to reduce the exponent by one. So, we need to multiply it by negative four 𝑧 plus nine all raised to the power of one-half minus one.

And we can simplify this expression. First, in our coefficient, we have one-half multiplied by negative four, which we can calculate as negative two. Next, in our exponent, we have one-half minus one, which is, of course, just equal to negative one-half. So, we can simplify our expression for d𝑦 by d𝑧 to be negative two times negative four 𝑧 plus nine all raised to the power of negative one-half.

In fact, we can simplify this further. By using our laws of exponents, instead of multiplying by a number raised to the power of negative one-half, we can divide by the square root of this number. So, we can rewrite this as negative two divided by the square root of negative four 𝑧 plus nine.

And there’s one more thing we’re going to do. Remember, we’re going to use the chain rule to find an expression for d𝑦 by dπ‘₯. After that, we’re going to substitute π‘₯ is equal to πœ‹ by eight 𝑧 into this expression. And to make this easier, we would want our expression for d𝑦 by d𝑧 to be in terms of π‘₯. To do this, we’re going to use our substitution 𝑧 is equal to the sec of six π‘₯. And by doing this, we get d𝑦 by d𝑧 is equal to negative two divided by the square root of negative four times the sec of six π‘₯ plus nine.

Now that we found an expression for d𝑦 by d𝑧, we need to find an expression for d𝑧 by dπ‘₯. That’s the derivative of the sec of six π‘₯ with respect to π‘₯. And to do this, we need to recall one of our standard rules for differentiating reciprocal trigonometric functions. For any real constant π‘Ž, the derivative of the sec π‘Žπ‘₯ with respect to π‘₯ is equal to π‘Ž times the tan of π‘Žπ‘₯ multiplied by the sec of π‘Žπ‘₯.

In this case, we can see our value of the constant π‘Ž is six. So, by setting π‘Ž equal to six, we get d𝑧 by dπ‘₯ is equal to six tan of six π‘₯ multiplied by the sec of six π‘₯. And now that we found the expressions for d𝑦 by d𝑧 and d𝑧 by dπ‘₯, we can substitute these into our chain rule to find d𝑦 by dπ‘₯. So, we multiply our expressions for d𝑦 by d𝑧 and d𝑧 by dπ‘₯ together. We get d𝑦 by dπ‘₯ is equal to negative two multiplied by six tan of six π‘₯ times the sec of six π‘₯ all divided by the square root of negative four times the sec of six π‘₯ plus nine.

And We can simplify this. In our numerator, we have two times six, which is equal to 12. But remember, the question is not just asking us to find an expression for d𝑦 by dπ‘₯. We need to find d𝑦 by dπ‘₯ at π‘₯ is equal to πœ‹ by 18. So, we need to substitute π‘₯ is equal to πœ‹ by 18 into this expression. And we could do this directly. However, we can simplify this. We notice all of our arguments are six π‘₯. And we can calculate six times πœ‹ by 18 is equal to πœ‹ by three. So, when we substitute π‘₯ is equal to πœ‹ by 18 into this expression, all of our arguments will be πœ‹ by three.

Substituting π‘₯ is equal to πœ‹ by 18 into this expression and simplifying, we get negative 12 times the tan of πœ‹ by three multiplied by the sec of πœ‹ by three all divided by the square root of negative four times the sec of πœ‹ by three plus nine. And now, we can just evaluate this expression. First, the tan of πœ‹ by three is equal to the square root of three. Next, we can see the sec of πœ‹ by three appears twice in this expression. And we know the sec of πœ‹ by three will be equal to one divided by the cos of πœ‹ by three. But the cos of πœ‹ by three is a half. And one divided by one-half is just equal to two.

So, we can replace both instances of the sec of πœ‹ by three with two. This gives us negative 12 times root three times two all divided by the square root of negative four times two plus nine. And we can just start calculating this expression. First, in our denominator, negative four times two is negative eight, plus nine is equal to one. So, our denominator is the square root of one, which is just equal to one. And in our numerator, we have 12 multiplied by two, which is equal to 24. So, we can just calculate this expression to give us negative 24 root three, which is our final answer.

Therefore, we were able to show if 𝑦 is equal to the square root of negative four 𝑧 plus nine and 𝑧 is equal to the sec of six π‘₯, then d𝑦 by dπ‘₯, when π‘₯ is equal to πœ‹ by 18, is equal to negative 24 root three.

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