Video: Finding Tangent Line Approximations

What is the tangent line approximation 𝑙(π‘₯) of √(1 βˆ’ π‘₯) near π‘₯ = 0?

02:49

Video Transcript

What is the tangent line approximation 𝑙 of π‘₯ of the square root of one minus π‘₯ near π‘₯ equals zero? Remember, if 𝑓 is differentiable at π‘Ž, then the equation for the tangent line approximation 𝑙 of π‘₯ is given by 𝑓 of π‘Ž plus 𝑓 prime of π‘Ž times π‘₯ minus π‘Ž. We’ll look at our example piece by piece. But first let’s find 𝑓 of π‘Ž. Our function 𝑓 of π‘₯ is the square root of one minus π‘₯. And we’re finding the tangent line approximation near π‘₯ equals zero. So we’re going to let π‘Ž be equal to zero. This means, in our expression, 𝑓 of π‘Ž is going to be 𝑓 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 we’re interested in is 𝑓 prime of π‘Ž. 𝑓 prime of π‘₯ is the derivative of 𝑓 with respect to π‘₯. So we’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 𝑦 is a function in 𝑒 and 𝑒 itself is a function in π‘₯, then d𝑦 by dπ‘₯ is the same as d𝑦 by d𝑒 times d𝑒 by dπ‘₯. If we say 𝑦 is the function the square root of one minus π‘₯, we can let 𝑒 be equal to one minus π‘₯ and 𝑦 be equal to the square root of 𝑒 which I’ve written as 𝑒 to the power of one-half.

d𝑒 by dπ‘₯, the derivative of one minus π‘₯ with respect to π‘₯, is simply negative one. And the derivative of 𝑦 with respect to 𝑒 is half times 𝑒 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 𝑒 to the negative a half multiplied by negative one. Replacing 𝑒 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 that’s just a special case of the chain rule.

So since we now know 𝑓 prime of π‘₯, we can evaluate 𝑓 prime of π‘Ž. 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 π‘₯ minus π‘Ž. And since π‘Ž is zero, this becomes π‘₯ minus zero which is just π‘₯.

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

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