### Video Transcript

In this video, weβll 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, weβre going to
investigate the application of the tangent line to the function. And how it allows us to approximate
more complicated functions. Weβll 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
π¦ equals π₯ 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 thereβs a brief interval to either
side for which the π¦ values along the tangent give a good approximation to the π¦
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 π of π₯.

Remember, the formula for the
equation of a straight line with a gradient of π and which passes through the
points π₯ one π¦ one is π¦ minus π¦ one equals π times π₯ minus π₯ one. We also know that the derivative of
a function π of π₯, π prime of π₯, 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 π of π₯ at a point π₯ equals π where π is differentiable
passes through the point π, π of π. And the gradient at this point will
be given by π prime of π.

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

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

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

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

So weβll begin by finding π prime
of π₯, thatβs the derivative of our function, and evaluating that at π₯ equals
negative two. The first derivative of our
function with respect to π₯ is three π₯ squared minus two π₯. 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, π₯
minus π? What do we know here? Well, we know that π is negative
two. So this becomes π₯ minus negative
two which is equal to π₯ plus two.

We can substitute each part into
our equation for π of π₯. And we get negative nine plus 16
times π₯ plus two. Distributing our expression, we get
negative nine plus 16π₯ minus 32. And then we simplify fully. And we see that the linear
approximation of the function π of π₯ equals π₯ cubed minus π₯ squared plus three
at π₯ equals negative two is π of π₯ equals 16π₯ 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. Letβs see what that might look
like.

What is the tangent line
approximation π of π₯ of the square root of one minus π₯ near π₯ equals zero? Remember, if π is differentiable
at π, then the equation for the tangent line approximation π of π₯ is given by π
of π plus π prime of π times π₯ minus π. Weβll look at our example piece by
piece. But first letβs find π of π. Our function π of π₯ is the square
root of one minus π₯. And weβre finding the tangent line
approximation near π₯ equals zero. So weβre going to let π be equal
to zero. This means, in our expression, π
of π is going to be π of zero. And we can evaluate this by
substituting π₯ 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 weβre interested in
is π prime of π. π prime of π₯ is the derivative of
π with respect to π₯. So weβre going to need to
differentiate the square root of one minus π₯ with respect to π₯. 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 π¦ is a function
in π’ and π’ itself is a function in π₯, then dπ¦ by dπ₯ is the same as dπ¦ by dπ’
times dπ’ by dπ₯. If we say π¦ is the function the
square root of one minus π₯, we can let π’ be equal to one minus π₯ and π¦ be equal
to the square root of π’ which Iβve written as π’ to the power of one-half.

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

So since we now know π prime of
π₯, we can evaluate π prime of π. Thatβs π prime of zero. So weβre going to substitute zero
into our formula for the derivative of our function. Itβs 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 weβre interested in is π₯ minus π. And since π is zero, this becomes
π₯ minus zero which is just π₯.

Substituting all of this into our
formula and we see the π of π₯ equals one plus negative a half times π₯. And this simplifies to one minus π₯
over two.

So weβve 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 π of π₯ equals π₯ sin of π₯ at π₯ equals two π.

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

Now, π prime of π is going to
require a little more work. Weβll find the derivative of our
function. Thatβs the derivative of π₯ sin π₯
with respect to π₯, noticing that we have a function which is itself the product of
two differentiable functions. Weβll, therefore, use the product
rule. This says that, for two
differentiable functions π’ and π£, the derivative of their product is π’ times dπ£
by dπ₯ plus π£ times dπ’ by dπ₯. For our function, weβll let π’ be
equal to π₯ and π£ be equal to sin π₯.

Weβre going to need to
differentiate each of these with respect to π₯. dπ’ by dπ₯ is one. And here, we recall the derivative
of sin π₯ with respect to π₯ is cos of π₯. And we substitute these into our
formula for the product rule. And we see that the derivative π
prime of π₯ is equal to π₯ times cos of π₯ plus sin π₯ times one. Thatβs π₯ cos π₯ plus sin π₯. To find π prime of two π, weβll
evaluate this when π₯ 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 π prime of two π is two π
times one plus zero which is simply two π.

Letβs substitute everything we now
have into the tangent line approximation formula. π of π is zero. π prime of π is two π. And π₯ minus π is π₯ minus two
π. We distribute our parentheses. And we see that the linear
approximation of our function π of π₯ equals π₯ sin π₯ at π₯ equals two π is two
ππ₯ minus four π squared.

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

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

Weβre told to use the linear
approximation of the function π of π₯ equals π₯ to the power of four. So we begin by finding the linear
approximation, sometimes called the tangent line approximation. This says that if π is
differentiable at some point π₯ equals π, then the equation that can be used to
find a linear approximation to the function at π₯ equals π is π of π plus π
prime of π times π₯ minus π. In this example, weβre 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,
weβre going to let π be equal to two. This means that π of π becomes π
of two. And we substitute π₯ equals two
into our function to get two to the power of four which is 16. Next, we find π prime of π. First, of course, weβre going to
need to find an expression for the derivative of π₯ to the power of four. So we begin by differentiating π₯
to the power of four with respect to π₯. And we get four π₯ cubed. This now means that π prime of π
becomes π 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 π of π₯ equals 16 plus
32 times π₯ minus two. And when we distribute our
parentheses we see π of π₯ is equal to 32π₯ minus 48. We can use this to approximate the
value of 1.999 to the power of four. We need to substitute π₯ 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, weβd get 15.96802399 and so on. So we can see that this is a very
good estimate. And thatβs 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 π of π₯ equals π to the power of π₯ at a suitable value of π₯,
estimate the value of π to the power of 0.1.

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

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

Weβll use this to approximate the
value of π to the power of 0.1 by finding π of 0.1. Thatβs 0.1 plus one which is
1.1. And an estimate to the value of π
to the 0.1 is 1.1. And if we type this into our
calculator, π to the 0.1 is 1.10517 and so on. Thatβs very close in value to our
estimation. And thatβs because 0.1 is fairly
close to zero. Had we tried the larger value, our
number might not have been so accurate. Letβs check that.

For example, π of 0.3 is 0.3 plus
one. So according to our approximation,
π to the 0.3 is approximately 1.3. Typing π to the 0.3 into our
calculator and we get 1.349858808, still not a bad approximation but not quite as
close as π to the power of 0.1.

In this video, weβve 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 π is differentiable at π₯
equals π, then the equation that can be used to find a linear approximation to the
function at this point is π of π₯ equals π of π plus π prime of π times π₯
minus π. We also saw that the closer the
value for π₯ is to the value of π, the more accurate the approximations will
be.