Question Video: Finding the Linear Approximation of a Trigonometric Function | Nagwa Question Video: Finding the Linear Approximation of a Trigonometric Function | Nagwa

Question Video: Finding the Linear Approximation of a Trigonometric Function Mathematics

Find the linear approximation of the function 𝑓(𝑥) = 𝑥 sin 𝑥 at 𝑥 = 2𝜋.

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Video Transcript

Find the linear approximation of the function 𝑓 of 𝑥 equals 𝑥 sin of 𝑥 at 𝑥 equals two 𝜋.

Remember, if 𝑓 is differentiable at 𝑥 equals 𝑎, then the equation for the tangent line approximation is given by 𝑙 of 𝑥 equals 𝑓 of 𝑎 plus 𝑓 prime of 𝑎 times 𝑥 minus 𝑎. In this example, we can let 𝑎 be equal to two 𝜋. We’re going to need to evaluate 𝑓 of 𝑎 and 𝑓 prime of 𝑎. Let’s begin with 𝑓 of 𝑎. In this case, that’s 𝑓 of two 𝜋. So we substitute 𝑥 is equal to two 𝜋 into 𝑥 sin 𝑥. And we get two 𝜋 times sin of two 𝜋. We should know that sin of two 𝜋 is equal to zero. So 𝑓 of two 𝜋 is two 𝜋 times zero which is just zero.

Now, 𝑓 prime of 𝑎 is going to require a little more work. We’ll find the derivative of our function. That’s the derivative of 𝑥 sin 𝑥 with respect to 𝑥, noticing that we have a function which is itself the product of two differentiable functions. We’ll, therefore, use the product rule. This says that, for two differentiable functions 𝑢 and 𝑣, the derivative of their product is 𝑢 times d𝑣 by d𝑥 plus 𝑣 times d𝑢 by d𝑥. For our function, we’ll let 𝑢 be equal to 𝑥 and 𝑣 be equal to sin 𝑥.

We’re going to need to differentiate each of these with respect to 𝑥. d𝑢 by d𝑥 is one. And here, we recall the derivative of sin 𝑥 with respect to 𝑥 is cos of 𝑥. And we substitute these into our formula for the product rule. And we see that the derivative 𝑓 prime of 𝑥 is equal to 𝑥 times cos of 𝑥 plus sin 𝑥 times one. That’s 𝑥 cos 𝑥 plus sin 𝑥. To find 𝑓 prime of two 𝜋, we’ll evaluate this when 𝑥 is equal to two 𝜋. That gives us two 𝜋 times cos of two 𝜋 plus sin of two 𝜋. We already said that sin of two 𝜋 is zero. And cos of two 𝜋 is one. So 𝑓 prime of two 𝜋 is two 𝜋 times one plus zero which is simply two 𝜋.

Let’s substitute everything we now have into the tangent line approximation formula. 𝑓 of 𝑎 is zero. 𝑓 prime of 𝑎 is two 𝜋. And 𝑥 minus 𝑎 is 𝑥 minus two 𝜋. We distribute our parentheses. And we see that the linear approximation of our function 𝑓 of 𝑥 equals 𝑥 sin 𝑥 at 𝑥 equals two 𝜋 is two 𝜋𝑥 minus four 𝜋 squared.

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