### Video Transcript

Determine the derivative of the function π of π₯ equals the square root of two π₯ minus 16 using the definition of the derivative.

Letβs firstly recall the definition of the derivative is π prime of π₯ equals the limit as β approaches zero of π of π₯ add β minus π of π₯ over β. Remember that to get π of π₯ add β, we just replace π₯ with π₯ add β in π of π₯. This gives us π of π₯ add β equals the square root of two multiplied by π₯ add β minus 16. We can then expand these parentheses to give us that π of π₯ plus β equals the square root of two π₯ plus two β minus 16.

Letβs now make this substitution into the definition of the derivative. Now, from here, thereβs actually a little bit of a trick that we use when weβre trying to find the derivative of a root function by definition. Itβs something we often do when weβre dealing with roots, and itβs to multiply by the conjugate. Remember, the conjugate is where we change the sign in the middle of the two terms. But we must also multiply the denominator by the same thing. So weβre essentially just multiplying what we had by one and, therefore, not technically changing anything.

Letβs start by multiplying the numerators together. We can think of this in the same way as if weβre to multiply two brackets containing algebraic terms. We can start by multiplying the square root of two π₯ add two β minus 16 with the square root of two π₯ add two β minus 16. To do this, we recall that the square root of π multiplied by the square root of π just gives us π. So multiplying these two terms together just gives us what we have inside the square root: two π₯ add two β minus 16. The same applies when we multiply negative the square root of two π₯ minus 16 with the square root of two π₯ minus 16. But this time, we have a negative at the front of one of the terms. So we need to be mindful of that. Using the same rules as before, this gives us negative two π₯ minus 16.

Now, we can also multiply negative the square root of two π₯ minus 16 with the square root of two π₯ add two β minus 16. We can just write these side by side to show that weβve multiplied them together. It doesnβt actually simplify any further than that. And finally, we need to multiply together the square root of two π₯ add two β minus 16 with the square root of two π₯ minus 16. When we do this, we see that these two terms actually cancel out. So our numerator is simply two π₯ add two β minus 16 minus two π₯ minus 16.

Now, letβs multiply the denominators together. We can just write this as β multiplied by the square root of two π₯ add two β minus 16 add the square root of two π₯ minus 16. Letβs simplify the numerator by expanding out this bracket, by multiplying it through by negative one. Notice that the minus 16 becomes a plus 16. From here, letβs simplify the numerator. Two π₯ cancels with the negative two π₯, and the negative 16 cancels with the add 16. So weβre just left with two β.

And now we notice that we can actually cancel something here in the numerator and the denominator. We have both β in the numerator and in the denominator. And since β is approaching zero and not equal to zero, this is okay to do. And because the β in the denominator is being multiplied by everything in the denominator, weβre okay to cancel through by this. So from here, letβs now apply the limit as β approaches zero. The only β we have is in the denominator. So weβre considering whatβs going to happen as this β approaches zero.

We can do this by direct substitution of β equals zero because this is a continuous function. This leaves us with two over the square root of two π₯ minus 16 add the square root of two π₯ minus 16. And because on the denominator we have the square root of two π₯ minus 16 add the square root of two π₯ minus 16, we can write this as two multiplied by the square root of two π₯ minus 16. And then we notice that the two in the numerator and the two in the denominator cancel each other. So weβre left with one over the square root of two π₯ minus 16. And thatβs our final answer.

One of the important things to note from this video is that when weβre finding the derivative of a root function using the definition of the derivative, weβre often going to need to multiply through by the conjugate. And from there, it was then just a case of doing some manipulation with algebra and applying the limit.