# Question Video: Finding Tangent Line Approximations Mathematics • Higher Education

What is the tangent line approximation π(π₯) of β(1 β π₯) near π₯ = 0?

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### 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. Weβll look at our example piece by piece. But first letβs find π of π. Our function π of π₯ is the square root of one minus π₯. And weβre finding the tangent line approximation near π₯ equals zero. So weβre 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 weβre interested in is π prime of π. π prime of π₯ is the derivative of π with respect to π₯. So weβre 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 Iβve 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 thatβs just a special case of the chain rule.

So since we now know π prime of π₯, we can evaluate π prime of π. Thatβs π prime of zero. So weβre going to substitute zero into our formula for the derivative of our function. Itβ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 weβre 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.