# Differentiate the function. Y= √x(x-8) ?

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To differentiate the function y = √x(x-8), we can use the product rule and the chain rule.

First, let's rewrite the function as y = x^(1/2) * (x-8)^(1/2).

Using the product rule, the derivative of y with respect to x is given by:

dy/dx = (x-8)^(1/2) * d/dx(x^(1/2)) + x^(1/2) * d/dx((x-8)^(1/2)).

Now, let's find the derivatives of each term using the chain rule:

d/dx(x^(1/2)) = (1/2) * x^(-1/2).

d/dx((x-8)^(1/2)) = (1/2) * (x-8)^(-1/2) * d/dx(x-8).

Since d/dx(x-8) = 1, the derivative simplifies to:

d/dx((x-8)^(1/2)) = (1/2) * (x-8)^(-1/2).

Substituting these derivatives back into the product rule equation, we have:

dy/dx = (x-8)^(1/2) * (1/2) * x^(-1/2) + x^(1/2) * (1/2) * (x-8)^(-1/2).

Simplifying further, we get:

dy/dx = (x-8)^(1/2) / (2√x) + x^(1/2) / (2√(x-8)).

Therefore, the derivative of y = √x(x-8) is:

dy/dx = (x-8)^(1/2) / (2√x) + x^(1/2) / (2√(x-8)).

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To differentiate the function ( y = \sqrt{x(x-8)} ), use the chain rule and the power rule:

- First, rewrite the function as ( y = (x(x-8))^{1/2} ).
- Apply the chain rule: ( y' = \frac{1}{2} (x(x-8))^{-1/2} \cdot (2x - 8) ).
- Simplify to get the derivative: ( y' = \frac{x - 4}{\sqrt{x(x-8)}} ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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