Question

Find the derivative of the function.

y= 1/2 [ x ((81-x^2)^(1/2)) + 81 arcsin (x/9)]

Answer 1

The first one.

Answer 2

\[y = \frac{ 1 }{ 2 } [ x \sqrt{81-x^2} + 81 \arcsin \frac{ x }{ 9 }\]

Answer 3

Apply product rule and chain rule.
What is the derivative of x?
What is the derivative of \(\sqrt{81-x^2} \)?
What is the derivative of \(\arcsin \frac{ x }{ 9 } \)?

Answer 4

x' = 1
derivative of \[\arcsin \frac{ x }{ 9 } = \frac{ 9 }{ \sqrt{1 - \frac{ x^2 }{ 81 }} }\]

Answer 5

@aum - The part with the chain rule is messing me up. :/

Answer 6

I know \[\sqrt{81-x^2} = (81-x^2)^{1/2}\]

Answer 7

Derivative of (something)^n = n * (something)^(n-1) * derivative of (something)
\[

\frac {d}{dx}\sqrt{81-x^2} = \frac {d}{dx}(81-x^2)^{1/2} = \frac 12~*~(81-x^2)^{-1/2}~*~\frac {d}{dx}(81-x^2) = ?

\]

Answer 8

\[\sqrt{\frac{ 2 }{ 81-x^2 }}\]

Answer 9

Forgot to multiply it by 2x...

Answer 10

The derivative in the last part of my previous reply is: (-2x). Multiply all three factors.

Answer 11

Oh, right. I had -2x on the paper I'm trying to work this out on.

Answer 12

\[
\frac {d}{dx}\sqrt{81-x^2} = \frac {-x}{\sqrt{81-x^2}}
\]

Answer 13

Similarly, derivative of arcsin(something) = 1/ sqrt(1 - (something)^2) * derivative of (something)
\[
\frac {d}{dx}\arcsin(x/9) = \frac {1}{\sqrt{1 - (x/9)^2}} ~*~ \frac {d}{dx}(x/9) = \\
\frac {1}{\sqrt{1 - (x/9)^2}} ~*~ \frac 19 =
\frac{9}{\sqrt{81-x^2}} * \frac 19 ~=~ \frac{1}{\sqrt{81-x^2}}
\]

Answer 14

Now you can put them all together and find y' using the product rule.

Answer 15

hello again
you got this ?