Video: Linear Approximation

In this video, we will use derivatives to find the equation of the line that approximates the function near a certain value and use differentials to approximate the change in the function.

16:15

Video Transcript

In this video, we’ll learn how we can use derivatives to find the equation of the line that approximates a function near a certain value. By this stage, you should feel confident finding the derivative of the function and the application of this. In this lesson, we’re going to investigate the application of the tangent line to the function. And how it allows us to approximate more complicated functions. We’ll then look at a number of examples of this application with varying degrees of difficulty and consider the geometrical interpretation.

Consider the tangent to the curve 𝑦 equals 𝑥 squared at the point with Cartesian coordinates one, one. We can see that the tangent lies close to the curve near the point of tangency, near one, one. If we zoom in on the graph and its tangent at the point of tangency, we see that there’s a brief interval to either side for which the 𝑦 values along the tangent give a good approximation to the 𝑦 values of our curve. And the more we magnify the graph near the point where it is differentiable, the flatter the graph appears and the more it will resemble its tangent. We can use this fact to develop a formula that can be used to give approximations for a function 𝑓 of 𝑥.

Remember, the formula for the equation of a straight line with a gradient of 𝑚 and which passes through the points 𝑥 one 𝑦 one is 𝑦 minus 𝑦 one equals 𝑚 times 𝑥 minus 𝑥 one. We also know that the derivative of a function 𝑓 of 𝑥, 𝑓 prime of 𝑥, tells us the gradient of the curve at a given point. In particular, it tells us the gradient of the tangent line to the curve at that point. We can say then that the tangent line to a function 𝑓 of 𝑥 at a point 𝑥 equals 𝑎 where 𝑓 is differentiable passes through the point 𝑎, 𝑓 of 𝑎. And the gradient at this point will be given by 𝑓 prime of 𝑎.

Substituting these values into our formula for the equation of a straight line and we get 𝑦 minus 𝑓 of 𝑎 equals 𝑓 prime of 𝑎 times 𝑥 minus 𝑎. We add 𝑓 of 𝑎 to both sides of our equation. And we get 𝑦 equals 𝑓 of 𝑎 plus 𝑓 prime of 𝑎 times 𝑥 minus 𝑎 as the equation for the tangent line. Let’s formalise this a bit. If 𝑓 is differentiable at 𝑥 equals 𝑎, then the equation of the tangent line 𝑙 of 𝑥 is given as 𝑓 of 𝑎 plus 𝑓 prime of 𝑎 times 𝑥 minus 𝑎. We call this the linear approximation to the function at 𝑥 equals 𝑎. Let’s have a look at the application of our definition.

Find the linear approximation of the function 𝑓 of 𝑥 equals 𝑥 cubed minus 𝑥 squared plus three at 𝑥 equals negative two.

Here, we have a polynomial function for which we’re being asked to find the linear approximation at 𝑥 equals negative two. Remember, if 𝑓 is differentiable at 𝑥 equals 𝑎, then the equation of the tangent line and the equation that can be used to find a linear approximation to the function at 𝑥 equals 𝑎 is given by 𝑙 of 𝑥 equals 𝑓 of 𝑎 plus 𝑓 prime of 𝑎 times 𝑥 minus 𝑎. Let’s break this down piece by piece.

In this example, we want to find the linear approximation at 𝑥 equals negative two. So we let 𝑎 equal negative two. Then 𝑓 of 𝑎 is 𝑓 of negative two. And we can find this value by substituting 𝑥 is equal to negative two into the function 𝑥 cubed minus 𝑥 squared plus three. 𝑓 of negative two is, therefore, negative two cubed minus negative two squared plus three which is negative nine. Next, we’re looking to find 𝑓 prime of 𝑎.

So we’ll begin by finding 𝑓 prime of 𝑥, that’s the derivative of our function, and evaluating that at 𝑥 equals negative two. The first derivative of our function with respect to 𝑥 is three 𝑥 squared minus two 𝑥. And this means the first derivative evaluated at negative two is given by three times negative two squared minus two times negative two which is 16. But what about this last part, 𝑥 minus 𝑎? What do we know here? Well, we know that 𝑎 is negative two. So this becomes 𝑥 minus negative two which is equal to 𝑥 plus two.

We can substitute each part into our equation for 𝑙 of 𝑥. And we get negative nine plus 16 times 𝑥 plus two. Distributing our expression, we get negative nine plus 16𝑥 minus 32. And then we simplify fully. And we see that the linear approximation of the function 𝑓 of 𝑥 equals 𝑥 cubed minus 𝑥 squared plus three at 𝑥 equals negative two is 𝑙 of 𝑥 equals 16𝑥 plus 23. This demonstrates a very simple example for how to find the linear approximation of a function. It is also important to realise that we can use this same process for more complicated functions by applying the rules for differentiation. Let’s see what that might look like.

What is the tangent line approximation 𝑙 of 𝑥 of the square root of one minus 𝑥 near 𝑥 equals zero? Remember, if 𝑓 is differentiable at 𝑎, then the equation for the tangent line approximation 𝑙 of 𝑥 is given by 𝑓 of 𝑎 plus 𝑓 prime of 𝑎 times 𝑥 minus 𝑎. We’ll look at our example piece by piece. But first let’s find 𝑓 of 𝑎. Our function 𝑓 of 𝑥 is the square root of one minus 𝑥. And we’re finding the tangent line approximation near 𝑥 equals zero. So we’re going to let 𝑎 be equal to zero. This means, in our expression, 𝑓 of 𝑎 is going to be 𝑓 of zero. And we can evaluate this by substituting 𝑥 equals zero into our function. And we get the square root of one minus zero or the square root of one which is simply one.

The next part we’re interested in is 𝑓 prime of 𝑎. 𝑓 prime of 𝑥 is the derivative of 𝑓 with respect to 𝑥. So we’re going to need to differentiate the square root of one minus 𝑥 with respect to 𝑥. We need to spot here that this is a function of a function or a composite function. And we can apply the chain rule. This says that if 𝑦 is a function in 𝑢 and 𝑢 itself is a function in 𝑥, then d𝑦 by d𝑥 is the same as d𝑦 by d𝑢 times d𝑢 by d𝑥. If we say 𝑦 is the function the square root of one minus 𝑥, we can let 𝑢 be equal to one minus 𝑥 and 𝑦 be equal to the square root of 𝑢 which I’ve written as 𝑢 to the power of one-half.

d𝑢 by d𝑥, the derivative of one minus 𝑥 with respect to 𝑥, is simply negative one. And the derivative of 𝑦 with respect to 𝑢 is half times 𝑢 to the power of one-half minus one which is negative one-half. So the derivative of the square root of one minus 𝑥 with respect to 𝑥 is a half times 𝑢 to the negative a half multiplied by negative one. Replacing 𝑢 with one minus 𝑥 and we see that the derivative of the square root of one minus 𝑥 with respect to 𝑥 is negative a half times one minus 𝑥 to the power of negative one-half. Note, at this stage, that we could have used the general power rule here. And that’s just a special case of the chain rule.

So since we now know 𝑓 prime of 𝑥, we can evaluate 𝑓 prime of 𝑎. That’s 𝑓 prime of zero. So we’re going to substitute zero into our formula for the derivative of our function. It’s negative a half times one minus zero to the power of negative a half which is negative one-half. The final part of our tangent line approximation that we’re interested in is 𝑥 minus 𝑎. And since 𝑎 is zero, this becomes 𝑥 minus zero which is just 𝑥.

Substituting all of this into our formula and we see the 𝑙 of 𝑥 equals one plus negative a half times 𝑥. And this simplifies to one minus 𝑥 over two. So we’ve seen so far how the chain rule can be used alongside the formula for the tangent line approximation. We can even use the tangent line approximation when dealing with trigonometric functions.

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. In our next two examples, we’ll look at how we can use the linear approximation for a function to approximate values.

By finding the linear approximation of the function 𝑓 of 𝑥 equals 𝑥 to the power of four at a suitable value of 𝑥, estimate the value of 1.999 to the power of four.

We’re told to use the linear approximation of the function 𝑓 of 𝑥 equals 𝑥 to the power of four. So we begin by finding the linear approximation, sometimes called the tangent line approximation. This says that if 𝑓 is differentiable at some point 𝑥 equals 𝑎, then the equation that can be used to find a linear approximation to the function at 𝑥 equals 𝑎 is 𝑓 of 𝑎 plus 𝑓 prime of 𝑎 times 𝑥 minus 𝑎. In this example, we’re going to try to approximate the value of 1.999 to the power of four. This is going to be very close in value to two to the power of four.

So in our linear approximation, we’re going to let 𝑎 be equal to two. This means that 𝑓 of 𝑎 becomes 𝑓 of two. And we substitute 𝑥 equals two into our function to get two to the power of four which is 16. Next, we find 𝑓 prime of 𝑎. First, of course, we’re going to need to find an expression for the derivative of 𝑥 to the power of four. So we begin by differentiating 𝑥 to the power of four with respect to 𝑥. And we get four 𝑥 cubed. This now means that 𝑓 prime of 𝑎 becomes 𝑓 prime of two which becomes four times two cubed. Two cubed is eight. So this is four times eight which is 32.

We substitute everything we now have into our tangent line approximation formula. And we get 𝑙 of 𝑥 equals 16 plus 32 times 𝑥 minus two. And when we distribute our parentheses we see 𝑙 of 𝑥 is equal to 32𝑥 minus 48. We can use this to approximate the value of 1.999 to the power of four. We need to substitute 𝑥 equals 1.999. And when we do, we get 32 times 1.999 minus 48 which is 15.968. An estimation for the value of 1.999 to the power of four is, therefore, 15.968. Now, if we were to type 1.999 to the power of four into our calculator, we’d get 15.96802399 and so on. So we can see that this is a very good estimate. And that’s because 1.999 is fairly close to two. How do we use this to evaluate, say, 2.3 to the power of four? Our answer might have been a little bit further out.

By finding the linear approximation of the function 𝑓 of 𝑥 equals 𝑒 to the power of 𝑥 at a suitable value of 𝑥, estimate the value of 𝑒 to the power of 0.1.

We’re told to use the linear approximation of the function 𝑓 of 𝑥 equals 𝑒 to the power of 𝑥. So we recall the formula. If 𝑓 is differentiable at 𝑥 equals 𝑎, then the equation that can be used to find the linear approximation to the function at 𝑥 equals 𝑎 is 𝑙 of 𝑥 equals 𝑓 of 𝑎 plus 𝑓 prime of 𝑎 times 𝑥 minus 𝑎. In this example, we’re trying to approximate the value of 𝑒 to the power of 0.1. This is going to be close to the value of 𝑒 to the power of zero. So we let 𝑎 be equal to zero. This means 𝑓 of 𝑎 is equal to 𝑓 of zero. And substituting zero into our function 𝑓 of 𝑥 equals 𝑒 to the power of 𝑥 gives 𝑒 to the power of zero which is one.

Next, we find 𝑓 prime of 𝑎. First, of course, we need to find an expression for the derivative of our function. So we different 𝑒 to the power of 𝑥 with respect to 𝑥. The first derivative of 𝑒 to the power of 𝑥 is 𝑒 to the power of 𝑥. So 𝑓 prime of 𝑎 becomes 𝑓 prime of zero which is 𝑒 to the power of zero. And once again, that’s one. Substituting what we know into our formula for our tangent line approximation and we see that 𝑙 of 𝑥 is equal to one plus one times 𝑥 minus zero. And that simplifies to 𝑥 plus one.

We’ll use this to approximate the value of 𝑒 to the power of 0.1 by finding 𝑙 of 0.1. That’s 0.1 plus one which is 1.1. And an estimate to the value of 𝑒 to the 0.1 is 1.1. And if we type this into our calculator, 𝑒 to the 0.1 is 1.10517 and so on. That’s very close in value to our estimation. And that’s because 0.1 is fairly close to zero. Had we tried the larger value, our number might not have been so accurate. Let’s check that.

For example, 𝑙 of 0.3 is 0.3 plus one. So according to our approximation, 𝑒 to the 0.3 is approximately 1.3. Typing 𝑒 to the 0.3 into our calculator and we get 1.349858808, still not a bad approximation but not quite as close as 𝑒 to the power of 0.1. In this video, we’ve learnt that we can use derivatives to find a tangent line approximation that can be used to approximate a function near a given value. If 𝑓 is differentiable at 𝑥 equals 𝑎, then the equation that can be used to find a linear approximation to the function at this point is 𝑙 of 𝑥 equals 𝑓 of 𝑎 plus 𝑓 prime of 𝑎 times 𝑥 minus 𝑎. We also saw that the closer the value for 𝑥 is to the value of 𝑎, the more accurate the approximations will be.

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