Video Transcript
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.
