Derivative of y = x sqrt(16-x^2)please show me steps if you can

You are trying to find the derivative of [tex]y = x \sqrt{16- x^{2} } [/tex].

First, put it into a format that's easier to read: 
[tex]y = x \sqrt{16- x^{2}} \\ y = x (16- x^{2})^{ \frac{1}{2} } [/tex]

Now you can see that you're trying to find the derivative of the product of two expressions: x and [tex] (16- x^{2})^{ \frac{1}{2}}[/tex].

Use the product rule of derivatives (see picture), where the derivative of y is equal to the derivative of the first factor, x, times the second factor, [tex](16- x^{2})^{ \frac{1}{2} }[/tex], plus the derivative of the second factor, [tex](16- x^{2})^{ \frac{1}{2}}[/tex], times the first factor.

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Start by finding the derivative of each factor:
1) The derivative of x is 1.
This is because of the power rule of derivatives. [tex] \frac{d}{dx} x^{n} = nx^{n-1} [/tex].
Remember x is the same thing as [tex]x^{1} [/tex]. Take the power (1) that the base (x) is raised to and multiply it by the base to get [tex]1*x^{1} [/tex]. Then subtract 1 from the power (1) that the base (x) is raised to: [tex] 1*x^{1-1} = 1*x^{0} = 1*1 = 1[/tex]. 1 is your derivative of x. 

2) The derivative of [tex](16- x^{2})^{ \frac{1}{2}} [/tex] is [tex]-x(16- x^{2})^{- \frac{1}{2}}[/tex].
This is because of the chain rule of derivatives. Find the derivative of the outside function (what's outside the parenthesis, then multiply that derivative by the inside function (what's inside the parenthesis).
For the outside function, pretend [tex](16- x^{2})^{ \frac{1}{2}}[/tex] is a basic [tex]x^{ \frac{1}{2}}[/tex], which makes it easier to understand and derive. That means x = [tex](16- x^{2})[/tex], but we'll worry about that later. Find the derivative of x^{ \frac{1}{2}}:
[tex] \frac{d}{dx}x^{ \frac{1}{2}} = \frac{1}{2}x^{ \frac{1}{2} -1} = \frac{1}{2}x^{-\frac{1}{2}}[/tex]

Now you can put x = [tex](16- x^{2})[/tex] back into x to get the derivative of the outside functon: [tex] \frac{1}{2}(16- x^{2})^{-\frac{1}{2}}[/tex].

That's only the first part of the chain rule. Now you have to find the derivative of the inside function (what's inside the parenthesis) and multiply it by that derivative of the outside function that we just found. We know that the inside function is [tex](16- x^{2})[/tex]. Find the derivative of [tex](16- x^{2})[/tex]:
[tex]\frac{d}{dx}(16- x^{2}) = \frac{d}{dx}16 - \frac{d}{dx}x^{2} = 0 - 2x = -2x[/tex]

That means the derivative of the inside function is -2x. Multiply that by the derivative of the outside function [tex] \frac{1}{2}(16- x^{2})^{-\frac{1}{2}}[/tex]:
[tex]\frac{1}{2} (16- x^{2})^{- \frac{1}{2}} \times (-2x) \\ = \frac{-2x}{2} (16- x^{2})^{- \frac{1}{2}}\\ = -x(16- x^{2})^{- \frac{1}{2}}[/tex]

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Now we go back to what we said earlier about the product rule of derivatives. The derivative of y (aka y') = (derivative of x)([tex](16- x^{2})^{ \frac{1}{2} }[/tex]) + (derivative of [tex](16- x^{2})^{ \frac{1}{2} }[/tex])(x)

We know from all our math above that:
Derivative of x = 1
Derivative of [tex](16- x^{2})^{ \frac{1}{2} }[/tex] =  [tex]-x(16- x^{2})^{- \frac{1}{2}}[/tex]

So follow the product rule to find y' (derivative of y):
[tex]y' = [1][(16- x^{2})^{ \frac{1}{2}}] + [-x(16- x^{2})^{- \frac{1}{2}}][x]\\ y' = (16- x^{2})^{ \frac{1}{2}} + -x^{2} (16- x^{2})^{- \frac{1}{2}}[/tex]

Finally, you can simplify [tex](16- x^{2})^{ \frac{1}{2}} + -x^{2} (16- x^{2})^{- \frac{1}{2}}[/tex]:
[tex](16- x^{2})^{ \frac{1}{2}} + -x^{2} (16- x^{2})^{- \frac{1}{2}}\\ = \sqrt{16- x^{2}} - \frac{ x^{2} }{\sqrt{16- x^{2}} } \\ = \frac{16- x^{2}}{\sqrt{16- x^{2}} } - \frac{ x^{2} }{\sqrt{16- x^{2}} }\\ = \frac{16- x^{2} - x^{2} }{\sqrt{16- x^{2}} }\\ = \frac{16- 2x^{2} }{\sqrt{16- x^{2}} }[/tex]

Your final answer is: [tex]\frac{16- 2x^{2} }{\sqrt{16- x^{2}} }[/tex].


Let me know if you're confused, and I'll try to explain the best I can :) It might be confusing at first, esp if you're new to derivatives. Hope it helped.