# Question Video: Finding the Derivative of a Root Function Using the Limit Definition of Derivatives Mathematics

Let π(π₯) = β6βπ₯ β 6. Use the definition of the derivative to determine πβ²(π₯).

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### Video Transcript

Let π of π₯ equal negative six times the square root of π₯ minus six. Use the definition of the derivative to determine π dash of π₯.

This π dash of π₯ or π prime of π₯, so called because the thing that looks like an apostrophe next to the π is called a dash or a prime, is the derivative of π of π₯. And in this question, we have to use the definition of the derivative. So letβs write it out.

By definition, the derivative of π of π₯, π dash of π₯, is the limit as β tends to zero of π of π₯ plus β minus π of π₯ all over β. Letβs attempt to apply this definition to our function. What is π of π₯ plus β? Well, π of π₯ is negative six times the square root of π₯ minus six.

So replacing π₯ by π₯ plus β, we see that π of π₯ plus β is negative six times the square root of π₯ plus β minus six. From this, we subtract π of π₯. And finally, we divide by β.

Letβs see if we can tidy up the numerator. We can get rid of these parentheses, which arenβt doing anything. And we can distribute the minus sign outside the second set of parentheses over the terms inside. Minus negative six square root π₯ becomes plus six square root π₯. And minus negative six becomes plus six.

We can then cancel a couple of terms. And we can also see a common factor of the two remaining terms, which we can factor out. We also took the opportunity to swap the two terms in the numerator. And finally, we can take the constant multiple of six outside the limit. This can be justified using one of our laws of limits.

Having performed all this algebraic manipulation, weβve simplified the limit that we have to find. Letβs clear some room and continue. Okay, so now that weβve simplified as far as we can, how do we go about evaluating this limit? The first thing to note is that direct substitution of β equals zero will give us the indeterminate form zero over zero.

Weβd like to somehow cancel out a common factor of β in the numerator and denominator so that when we substitute in, this doesnβt happen. The trick to doing this is to multiply both numerator and denominator by the conjugate of our numerator.

You get the conjugate of an expression with two terms by flipping the sign of the second term. So the conjugate of π plus π is π minus π. And the conjugate of π minus π is π plus π. We therefore multiply it by square root π₯ plus square root π₯ plus β.

We can simplify the numerator of our fraction using the distributive property or FOIL. And doing so, we notice that the two cross-terms in the middle cancel and the other two terms we can simplify. The square root of π₯ squared is π₯. And the square root of π₯ plus β squared is π₯ plus β. Thereβs nothing we can really do with the denominator. So we leave it as it is.

Letβs tidy up the numerator and see what weβre left with. The numerator is simply π₯ minus π₯ plus β. Distributing the minus sign over the parentheses, we get π₯ minus π₯ minus β. And the π₯ terms cancel. So the numerator is simply negative β.

We have succeeded in getting a common factor of β in the numerator and denominator, which we can now cancel. Upon canceling this common factor of β, we see that we can now substitute directly β equals zero. Doing this, being careful to note that the numerator is now negative one, we get the following expression.

The square root of π₯ plus zero is just the square root of π₯. And we can combine this with the other square root π₯ to get two square root π₯. And simplifying this by combining the constants, we get negative three over square root π₯.

Remember that this is a derivative, π dash of π₯, of the function in the question. So if you input a value of π₯ to this derivative function, π dash of π₯, you get out the slope of the tangent, the original function, π of π₯, at this value of π₯.