Video: Finding Tangent Line Approximations

Find the local linearization of 1/√(1 + π‘₯) near π‘₯ = 0.

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Video Transcript

Find the local linearization of one divided by the square root of one plus π‘₯ near π‘₯ is equal to zero.

The question wants us to find the local linearization of one divided by the square root of one plus π‘₯ near the point where π‘₯ is equal to zero. What this means is we need to find tangent to the curve 𝑦 is equal to one divided by the square root of one plus π‘₯ when π‘₯ is equal to zero. We can do this by recalling if our function 𝑓 is differentiable at the point where π‘₯ is equal to π‘Ž, then the tangent to the curve 𝑦 is equal to 𝑓 of π‘₯ when π‘₯ is equal to π‘Ž is given by 𝐿 of π‘₯ is equal to 𝑓 evaluated at π‘Ž plus the derivative of 𝑓 evaluated at π‘Ž multiplied by π‘₯ minus π‘Ž. So, we want to set our function 𝑓 to be equal to one divided by the square root of one plus π‘₯, and we want to set π‘Ž equal to zero.

Then, if we can show that one divided by the square root of one plus π‘₯ is differentiable at the point where π‘₯ is equal to zero, then our local linearization near π‘₯ is equal to zero is given by 𝑓 evaluated at zero plus 𝑓 prime evaluated at zero multiplied by π‘₯ minus zero. To help us calculate our derivative function, we can notice that one divided by the square root of one plus π‘₯ can be rewritten as one plus π‘₯ to the power of negative a half. However, we still can’t differentiate this directly. We need to notice that one plus π‘₯ all raised to the power of negative a half is the composition of two functions.

If we set 𝑔 of π‘₯ to be equal to π‘₯ to the power of negative a half and β„Ž of π‘₯ to be equal to one plus π‘₯, then we can notice that 𝑓 of π‘₯ is the composition of 𝑔 with β„Ž. We’re now ready to calculate our derivative function. We’ll do this by using the chain rule. We have that if the function 𝑓 of π‘₯ is equal to the composition of 𝑔 and β„Ž, then 𝑓 prime of π‘₯ is equal to β„Ž prime of π‘₯ multiplied by 𝑔 prime evaluated at β„Ž of π‘₯. So, by using the chain rule, we have our derivative function, 𝑓 prime of π‘₯, is equal to β„Ž prime of π‘₯ multiplied by 𝑔 prime evaluated at β„Ž of π‘₯. We can calculate 𝑔 prime of π‘₯ by using our power rule for differentiation. We multiply by the exponent and then reduce this exponent by one. This gives us negative a half multiplied by π‘₯ to the power of negative three over two.

Similarly, we can calculate β„Ž prime of π‘₯. The derivative of the constant one is just equal to zero, and the derivative of π‘₯ is equal to one. So, β„Ž prime of π‘₯ can be simplified to be equal to one. So, 𝑓 prime of π‘₯ is equal to β„Ž prime of π‘₯ which is one. And we multiply this by 𝑔 prime of π‘₯. But we want to evaluate this at the point β„Ž of π‘₯. So, instead of writing π‘₯, we write β„Ž of π‘₯ which is one plus π‘₯. This gives us one multiplied by negative a half multiplied by one plus π‘₯ to the power of negative three over two. Simplifying this gives us that 𝑓 prime of π‘₯ is equal to negative one divided by two multiplied by one plus π‘₯ to the power of three over two.

We can notice that our derivative function, 𝑓 prime of π‘₯, is the quotient of two continuous functions. Therefore, in particular, it must be continuous when the denominator is not equal to zero. Therefore, we’ve shown 𝑓 prime is continuous at the point where π‘₯ is equal to zero. So, our function 𝑓 is differentiable at the point where π‘₯ is equal to zero. So, we’re now ready to calculate our local linearization about π‘₯ is equal to zero. We’ll start by calculating 𝑓 prime evaluated at zero. This is equal to negative one divided by two multiplied by one plus zero to the power of three over two. We have that one plus zero is equal to one. And then raising this to the power three over two just gives us one.

So, we have negative one divided by two multiplied by one which is just equal to negative a half. So, our linearization is equal to 𝑓 evaluated at zero plus 𝑓 prime evaluated at zero multiplied by π‘₯ minus zero. We have 𝑓 evaluated at zero is one divided by the square root of one plus zero. We calculated 𝑓 prime evaluated at zero to be negative a half. And π‘₯ minus zero is just equal to π‘₯. Simplifying this gives us one minus π‘₯ over two. Therefore, we have shown that the local linearization of one divided by the square root of one plus π‘₯ near the point where π‘₯ is equal to zero is one minus π‘₯ over two.

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