### Video Transcript

Determine the derivative of the function ๐ of ๐ฅ equals the square root of two ๐ฅ minus 16 using the definition of the derivative.

Letโs firstly recall the definition of the derivative is ๐ prime of ๐ฅ equals the limit as โ approaches zero of ๐ of ๐ฅ add โ minus ๐ of ๐ฅ over โ. Remember that to get ๐ of ๐ฅ add โ, we just replace ๐ฅ with ๐ฅ add โ in ๐ of ๐ฅ. This gives us ๐ of ๐ฅ add โ equals the square root of two multiplied by ๐ฅ add โ minus 16. We can then expand these parentheses to give us that ๐ of ๐ฅ plus โ equals the square root of two ๐ฅ plus two โ minus 16.

Letโs now make this substitution into the definition of the derivative. Now, from here, thereโs actually a little bit of a trick that we use when weโre trying to find the derivative of a root function by definition. Itโs something we often do when weโre dealing with roots, and itโs 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 weโre essentially just multiplying what we had by one and, therefore, not technically changing anything.

Letโs start by multiplying the numerators together. We can think of this in the same way as if weโre to multiply two brackets containing algebraic terms. We can start by multiplying the square root of two ๐ฅ add two โ minus 16 with the square root of two ๐ฅ add two โ minus 16. To do this, we recall that the square root of ๐ multiplied by the square root of ๐ just gives us ๐. So multiplying these two terms together just gives us what we have inside the square root: two ๐ฅ add two โ minus 16. The same applies when we multiply negative the square root of two ๐ฅ minus 16 with the square root of two ๐ฅ 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 ๐ฅ minus 16.

Now, we can also multiply negative the square root of two ๐ฅ minus 16 with the square root of two ๐ฅ add two โ minus 16. We can just write these side by side to show that weโve multiplied them together. It doesnโt actually simplify any further than that. And finally, we need to multiply together the square root of two ๐ฅ add two โ minus 16 with the square root of two ๐ฅ minus 16. When we do this, we see that these two terms actually cancel out. So our numerator is simply two ๐ฅ add two โ minus 16 minus two ๐ฅ minus 16.

Now, letโs multiply the denominators together. We can just write this as โ multiplied by the square root of two ๐ฅ add two โ minus 16 add the square root of two ๐ฅ minus 16. Letโs 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, letโs simplify the numerator. Two ๐ฅ cancels with the negative two ๐ฅ, and the negative 16 cancels with the add 16. So weโre just left with two โ.

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

We can do this by direct substitution of โ equals zero because this is a continuous function. This leaves us with two over the square root of two ๐ฅ minus 16 add the square root of two ๐ฅ minus 16. And because on the denominator we have the square root of two ๐ฅ minus 16 add the square root of two ๐ฅ minus 16, we can write this as two multiplied by the square root of two ๐ฅ minus 16. And then we notice that the two in the numerator and the two in the denominator cancel each other. So weโre left with one over the square root of two ๐ฅ minus 16. And thatโs our final answer.

One of the important things to note from this video is that when weโre finding the derivative of a root function using the definition of the derivative, weโre 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.