Question

derivative of x(8-x^2)^(1/2)

Answer 1

use Product Rule

Answer 2

Your function is:\[y=x \sqrt{8-x^2}=x \cdot(8-x^2)^{\frac{ 1 }{ 2 }}\]
To find the derivative, you have to use the product rule.

Answer 3

In this case this means:\[y'=x'\cdot(8-x^2)^{\frac{ 1 }{ 2 }}+x \cdot ((8-x^2)^{\frac{ 1 }{ 2 }})'\]First part is easy, because x' = 1.
Second part requires more work: you'll have to use the rule:\[(x^n)' =nx^{n-1}\]and while doing this, you realize that you also need the chain rule: instead of x there is 8-x² between the brackets.

Answer 4

So\[y'=1 \cdot (8-x^2)^{\frac{ 1 }{ 2 }} + x \cdot \frac{ 1 }{ 2 }(8-x^2)^{-\frac{ 1 }{ 2 }} \cdot -2x\]Although this is the derivative, extensive cleaning is needed ;)
First, write again with radicals:\[y'=\sqrt{8-x^2}-\frac{ x^2 }{ \sqrt{8-x^2} }\]
Second, Wrtie as one fraction:\[y'=\sqrt{8-x^2}-\frac{ x^2 }{ \sqrt{8-x^2} }=\sqrt{8-x^2} \cdot \frac{ \sqrt{8-x^2} }{ \sqrt{8-x^2} }-\frac{ x^2 }{ \sqrt{8-x^2} }=\]\[y'=\frac{ 8-x^2 }{ \sqrt{8-x^2} }-\frac{ x^2 }{ \sqrt{8-x^2}}=\frac{ 8-2x^2 }{ \sqrt{8-x^2} }\]
If you write it this way, it is easy to solve y'=0, which is needed if you want to know about the extreme values of this function.

So the hard part of this is for many people not deriving the function, but simplifying afterwards...