### Video Transcript

Estimate the function π of π₯ is equal to two π₯ times the natural logarithm of three π₯ with a third-degree Taylor polynomial at π₯ is equal to one.

The question wants us to find a third-degree Taylor polynomial approximation centered at π₯ is equal to one for our function π of π₯ is equal to two π₯ times the natural logarithm of three π₯. We recall we can find an πth-degree Taylor polynomial approximation of the function π of π₯ at π₯ is equal to π as the following polynomial. π π of π₯ is equal to π evaluated at π plus π prime evaluated at π divided by one factorial multiplied by π₯ minus π. Plus the second derivative of π evaluated at π divided by two factorial multiplied by π₯ minus π squared. And we keep adding terms of this form all the way up to the πth derivative of π evaluated at π divided by π factorial multiplied by π₯ minus π to the πth power.

We want to find the third-degree Taylor polynomial approximation of our function, so weβll set π equal to three. And we want our Taylor polynomial to be centered at π₯ is equal to one. So, weβll set π equal to one. We see to find the third-degree Taylor polynomial approximation, we need to find the first three derivatives of our function π of π₯ evaluated at π₯ is equal to π. So, we need to find the first three derivatives of our function π of π₯. Weβll start by finding the first derivative, π prime of π₯.

To do this, we notice that our function π of π₯ is the product of two functions. To help us evaluate this derivative, we recall the product rule for differentiation, which tells us the derivative of the product of two functions π’ and π£ with respect to π₯ is equal to π’ prime π£ plus π’π£ prime. So, to differentiate π of π₯, we start by differentiating two π₯ with respect to π₯. Thatβs equal to two. We then multiply this by the natural logarithms of three π₯.

Now, we need to add two π₯ multiplied by the derivative of the natural logarithm of three π₯. To help us differentiate this, we recall the product rule for logarithms, which tells us the log of ππ₯ is equal to the log of π plus the log of π₯. This means we can rewrite the natural logarithm of three π₯ as the natural logarithm of three plus the natural logarithm of π₯. But the natural logarithm of three is a constant. So, when we differentiate this, we just get zero.

And we know the derivative of the natural logarithm of π₯ with respect to π₯ is equal to one divided by π₯. So, the derivative of the natural logarithm of three π₯ is equal to one divided by π₯. And we can simplify two π₯ multiplied by one over π₯ to just be equal to two. So, weβve shown that π prime of π₯ is equal to two times the natural logarithm of three π₯ plus two. To find our second derivative of π with respect to π₯, we need to differentiate the first derivative of π of π₯ with respect to π₯.

We already showed that the derivative of the natural logarithm of three π₯ is just one divided by π₯. So, the derivative of two times the natural logarithm of three π₯ is just equal to two times one over π₯. And of course, the derivative of the constant two is just equal to zero. Weβll write this as two times π₯ to the power of negative one.

Finally, to find the third derivative of π of π₯ with respect to π₯, we need to differentiate the second derivative of π of π₯ with respect to π₯. We can do this by using the power rule for differentiation. We multiply by the exponent of negative one and then reduce the exponent by one. This gives us negative two multiplied by π₯ to the power of negative two. Since weβve now found the first three derivatives of our function π of π₯, we now need to evaluate all four of these functions at the center of our approximation, π₯ is equal to one.

π evaluated at one is equal to two times one times the natural logarithm of three times one, which simplifies to give us two times the natural logarithm of three. π prime evaluated at one is equal to two times the natural logarithm of three times one plus two, which simplifies to give us two times the natural logarithm of three plus two.

The second derivative of π evaluated at π₯ is equal to one is two times the reciprocal of one. And we know the reciprocal of one is just equal to one. So, this simplifies to give us two. Finally, the third derivative of π evaluated at π₯ is equal to one is negative two times one to the power of negative two. And one to the power of negative two is one divided by one squared, which is just equal to one. So, this simplifies to give us negative two.

Weβre now ready to find the third-degree Taylor polynomial of our function π of π₯ at π₯ is equal to one. Itβs given by the polynomial π evaluated at one plus π prime evaluated at one over one factorial times π₯ minus one. Plus π double prime evaluated at one over two factorial times π₯ minus one squared. Plus the third derivative of π evaluated at one over three factorial times π₯ minus one cubed.

Weβve already calculated π, π prime, π double prime, and π triple prime evaluated at π₯ is equal to one. Substituting these four values into our expression gives us two times the natural logarithm of three. Plus two times the natural logarithm of three. Plus two over one factorial times π₯ minus one. Plus two over two factorial times π₯ minus one squared. Plus negative two divided by three factorial times π₯ minus one cubed.

And we can simplify this. We know one factorial is just equal to one, so dividing by this doesnβt change anything. Two factorial is equal to two, so two divided by two factorial is just equal to one. And three factorial is equal to six, so negative two divided by three factorial is equal to negative one-third. Therefore, weβve shown that the third-degree Taylor polynomial approximation at π₯ is equal to one of our function π of π₯ is equal to two π₯ times the natural logarithm of three π₯. Is equal to two times the natural logarithm of three plus two times the natural algorithm of three plus two times π₯ minus one plus π₯ minus one squared minus one-third π₯ minus one cubed.