In this problem, we're asked to find the derivative of \( f(x) = \tan(3 - \sin(2x)) \). Now, this is a little different than some of the problems that we have been working with because we don't have anything raised to a power. However, we are still working with multiple functions here, and we are going to have to apply the chain rule multiple times. Let's go ahead and break this down. Looking at this function, my outermost function is the tangent.
Now working my way inside, I see that inside the argument of tangent, I have the function \( 3 - \sin(2x) \). So, what is my innermost function? Well, I see that inside the argument of sine, I have \( 2x \). So that is actually my innermost function. It might be a little clearer here if I add an extra set of parentheses around that \( 2x \) showing that this is my innermost function.
Let's go ahead and use our chain rule here to find our derivative \( f'(x) \). Remember that when using the chain rule, we are always starting from the outside and working our way in. Here, our outermost function is the tangent. So what is the derivative of tangent? The derivative of the tangent is \( \sec^2 \).
Since we're using the chain rule, whatever is inside the function remains unchanged. So my argument is still \( 3 - \sin(2x) \). This is where the argument for the first derivative ends. Working our way further inside, I consider the next function, \( 3 - \sin(2x) \).
The derivative of \( 3 \) is just \( 0 \), so I don’t have to worry about that. There is also a negative sign. The derivative of \( \sin(2x) \) is \( \cos(2x) \). Remember, whatever is in parentheses or within that function remains the same. One final step here, as we need to account for that innermost function \( 2x \) by taking its derivative and multiplying by it. The derivative of \( 2x \) is just \( 2 \). So, my last step here is to multiply by that final derivative of \( 2 \). Now we can simplify this a little bit and rewrite it to make it look a bit prettier.
Our derivative function, \( f'(x) \), pulling that constant up to the front and also moving this cosine to the front. The expression is \( -2 \cos(2x) \cdot \sec^2(3 - \sin(2x)) \), using the same argument that we started with here. If you wrote this in a different order, that's totally fine. It is still an equivalent statement.
This is our final derivative, and I will see you in the next video.