Question Video: Differentiating Trigonometric Functions Whose Arguments Are Linear Functions Mathematics • Higher Education

If 𝑦 = 3 cos (8π‘₯ βˆ’ 3), find d𝑦/dπ‘₯.

06:20

Video Transcript

If 𝑦 is equal to three times the cos of eight π‘₯ minus three, find d𝑦 by dπ‘₯.

We’re given a function 𝑦 which is a trigonometric function. And we’re asked to find d𝑦 by dπ‘₯. That’s the derivative of 𝑦 with respect to π‘₯. However, we could see a problem. Our argument inside of our trigonometric function is not just a multiple of π‘₯; it’s eight π‘₯ minus three. And we don’t know how to differentiate trigonometric functions in this form. There’s a few different ways we could approach this problem. For example, we could try and rewrite this in a form which we can differentiate. And to do this, we can recall the angle addition or subtraction formula for cos.

We know the cos of π‘Ž plus or minus 𝑏 is equal to the cos of π‘Ž times the cos of 𝑏 minus, plus the sin of π‘Ž times the sin of 𝑏. In our case, we want to take the difference of two angles. So we will write this in the following way. Then by setting π‘Ž equal to eight π‘₯ and 𝑏 equal to three, we get the cos of eight π‘₯ minus three is equal to the cos of eight π‘₯ times the cos of three plus the sin of eight π‘₯ times the sin of three. We can then multiply both sides of this equation through by three. And the left-hand side of this equation is exactly equal to 𝑦. And we know how to differentiate the right-hand side of this equation.

Differentiating both sides of this equation with respect to π‘₯, we get d𝑦 by dπ‘₯ is equal to the derivative of three cos of eight π‘₯ times the cos of three plus three times the sin of eight π‘₯ times the sin of three with respect to π‘₯. And to differentiate this, we just need to notice the cos of three and the sin of three are constants. So the only parts varying with respect to π‘₯ are the cos of eight π‘₯ and the sin of eight π‘₯. So we’ll evaluate this derivative term by term.

To differentiate our first term, we recall for any real constant π‘˜, the derivative of the cos of π‘˜π‘₯ with respect to π‘₯ is equal to negative π‘˜ times the sin of π‘˜π‘₯. In our first term, the value of π‘˜ is equal to eight. So by applying this, we can differentiate our first term with respect to π‘₯ to get three times negative eight sin of eight π‘₯ times the cos of three. And we can simplify our coefficient since three times negative eight is equal to negative 24.

And to differentiate our second term, we need to recall another one of our trigonometric derivative results. For any real constant π‘˜, the derivative of the sin of π‘˜π‘₯ with respect to π‘₯ is equal to π‘˜ times the cos of π‘˜π‘₯. And in our second term, we can see the value of π‘˜ is equal to eight. So by applying this rule with the value of π‘˜ equal to eight, we can differentiate our second term to get three times eight cos of eight π‘₯ multiplied by the sin of three. And once again, we can simplify our coefficient three times eight is equal to 24. And this gives us an expression for d𝑦 by dπ‘₯ just like the question asked.

However, this expression is very messy. We can actually simplify this by using another one of our angle rules. We need to recall that the sin of π‘Ž minus 𝑏 is equal to the sin of π‘Ž times the cos of 𝑏 minus the cos of π‘Ž times the sin of 𝑏. And it might be difficult to see how we’re going to use this simplify expression. So we’ll rewrite our expression slightly to make this easier. We’ll take out a factor of negative 24. And if we do this, we get the following expression. And we can see this is exactly what we have with our value of π‘Ž said to be eight π‘₯ and our value of 𝑏 said to be three.

So we can replace everything inside of our parentheses with the sin of eight π‘₯ minus three. This gives us that d𝑦 by dπ‘₯ is equal to negative 24 times the sin of eight π‘₯ minus three. However, this is just one way we could have answered this question. We’ll go through another way using a more graphical approach. To do this, let’s take a look at our function 𝑦 is equal to three times the cos of eight π‘₯ minus three. Now, we don’t know how to differentiate this directly. However, we do know how to differentiate similar functions. For example, we know how to differentiate three times the cos of eight π‘₯.

We might then be tempted to rewrite our function in the following form: Three times the cos of eight times π‘₯ minus three over eight. And if we did do this, we would notice something interesting. This is just a horizontal transformation of the curve three times the cos of eight π‘₯. We’re just transforming it horizontally to the right three over eight units. And to see why this is useful, consider the following sketch of 𝑦 is equal to three times the cos of eight π‘₯. This is a curve which we know how to differentiate. And remember, for any point, the derivative of the curve at that point will be the slope of its tangent line at that point.

But think what would happen if we were to translate this entire graph to the right three over eight units? If each point on our curve is moving to the right three over eight units, then the slope of our tangent lines will be the same, just three over eight units to the right. So instead of finding d𝑦 by dπ‘₯ directly, we can just differentiate 𝑦 is equal to three times the cos of eight π‘₯ and then translate our answer afterwards.

So we want to differentiate three times the cos of eight π‘₯ with respect to π‘₯. In fact, we’ve already done this and written the rule we used to do this. We set our value of π‘˜ equal to eight. So differentiating this, we get three times negative eight sin of eight π‘₯ which we can simplify to give us negative 24 times the sin of eight π‘₯. But remember, this is giving us the slope to the tangent lines to the curve 𝑦 is equal to three cos of eight π‘₯. So we need to translate this three over eight units to the right. And to do this, we need to subtract negative three over eight from our value of π‘₯. So by translating our slope graph three over eight units to the right, we get d𝑦 by dπ‘₯ is equal to negative 24 times the sin of eight multiplied by π‘₯ minus three over eight.

And of course, we can simplify this since eight times π‘₯ minus three over eight is equal to eight π‘₯ minus three. So this gives us two different methods of showing that d𝑦 by dπ‘₯ is equal to negative 24 times the sin of eight π‘₯ minus three. And both of these methods were a bit difficult and complicated to use. In fact, there’s a method outside of the scope of this video which will make these types of problems easier. It’s called the chain rule.

This says, if 𝑓 and 𝑔 are differentiable functions, then the derivative of 𝑓 composed with 𝑔 of π‘₯ with respect to π‘₯ is equal to 𝑔 prime of π‘₯ times 𝑓 prime evaluated at 𝑔 of π‘₯. And in fact, we can use the chain rule or either of the two methods we’ve just discussed to prove two very useful results for differentiating trigonometric functions. However, for now, we were able to show if 𝑦 is equal to three cos of eight π‘₯ minus three, then d𝑦 by dπ‘₯ is equal to negative 24 sin of eight π‘₯ minus three.

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