Finding the derivative

[Sga001]

Hello, i have a math problem that i tried, and i couldn't do it (at least i did not get the right answer).

My answer: 10/root(16-x^2) Real Answer: -x^2/ root (16-x^2)

Problem: take the derivatve of the function:

\(\displaystyle \
y = 8\arcsin (\frac{x}{4}) - x\frac{{\sqrt {16 - x^2 } }}{2}
\\)

I could copy what i have done but its a little more than 1 page, and to type it into TeX lang will take me a long time (since im not really experienced).


Thank You

 

[soroban]

Hello, Sga001!

This is a dangerous problem, filled with quicksand and landmines . . .

Differentiate: \(\displaystyle \L\:y\:=\:8\cdot\arcsin\left(\frac{x}{4}\right) \,-\,x\frac{\sqrt{16\,-\,x^2}}{2}\)

The given answer is wrong . . . there's no "minus"

We have: \(\displaystyle \L\:y \;=\;8\cdot\arcsin\left(\frac{x}{4}\right)\,-\,\frac{1}{2}\cdot x\cdot(16\,-\,x^2}^{\frac{1}{2}}\)

Then: \(\displaystyle \L\:y' \;=\;8\cdot\frac{\frac{1}{4}}{\sqrt{1 - (\frac{x}{4})^2}} \,-\,\frac{1}{2}\cdot\left[x\cdot\frac{1}{2}\cdot(16\,-\,x^2)^{-\frac{1}{2}}\cdot(-2x)\,+\,1\cdot(16\,-\,x^2)^{\frac{1}{2}}\right]\)

. . \(\displaystyle \L y' \;=\;\frac{2}{\sqrt{1\,-\,\frac{x^2}{16}}} \,-\,\frac{1}{2}\cdot\left[\frac{-x}{\sqrt{16\,-\,x^2}}\,+\,\sqrt{16\,-\,x^2}\right]\)

. . \(\displaystyle \L y' \;=\;\frac{2}{\sqrt{\frac{16\,-\,x^2}{16}}} \,-\, \frac{1}{2}\cdot\left[\frac{-x^2\,+\,16\,-\,x^2}{\sqrt{16\,-\,x^2}}\right] \;=\;\frac{2}{\frac{\sqrt{16\,-\,x^2}}{4}} \,-\,\frac{1}{2}\cdot\frac{16\,-\,2x^2}{\sqrt{16\,-\,x^2}}\)

. . \(\displaystyle \L y' \;= \;\frac{8}{\sqrt{16\,-\,x^2}} \,-\,\frac{8\,-\,x^2}{\sqrt{16\,-\,x^2}} \;=\;\frac{8\,-\,(8\,-\,x^2)}{\sqrt{16\,-\,x^2}} \;=\;\fbox{\frac{x^2}{\sqrt{16\,-\,x^2}}}\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I know my answer is correct . . . I checked it by integration.
(Kids, don't try this at home. .I'm a trained professional.)

 

[galactus]

It's just algebra troubles, no doubt.

I will skip all the calculus and get right to the algebra. OK?.

\(\displaystyle \L\\\frac{d}{dx}[8sin^{-1}(\frac{x}{4})]=\frac{8}{\sqrt{16-x^{2}}\)

\(\displaystyle \L\\\frac{d}{dx}[\frac{x\sqrt{16-x^{2}}}{2}]=\frac{\sqrt{16-x^{2}}}{2}-\frac{x^{2}}{2\sqrt{16-x^{2}}}\)

So, we have:

\(\displaystyle \L\\\frac{8}{\sqrt{16-x^{2}}}-\left(\frac{\sqrt{16-x^{2}}}{2}-\frac{x^{2}}{2\sqrt{16-x^{2}}}\right)\)

Let's work inside the parentheses first:

\(\displaystyle \L\\\frac{\sqrt{16-x^{2}}}{2}\cdot\frac{\sqrt{16-x^{2}}}{\sqrt{16-x^{2}}}-\frac{x^{2}}{2\sqrt{16-x^{2}}}\)

=\(\displaystyle \L\\\frac{16-x^{2}}{2\sqrt{16-x^{2}}}-\frac{x^{2}}{2\sqrt{16-x^{2}}}\)

=\(\displaystyle \L\\\frac{16-2x^{2}}{2\sqrt{16-x^{2}}}\)

Therefore, we have:

\(\displaystyle \L\\\frac{8}{\sqrt{16-x^{2}}}-\left(\frac{16-2x^{2}}{2\sqrt{16-x^{2}}}\right)\)

\(\displaystyle \L\\\frac{2}{2}\cdot\frac{8}{\sqrt{16-x^{2}}}-\left(\frac{16-2x^{2}}{2\sqrt{16-x^{2}}}\right)\)

=\(\displaystyle \L\\\frac{16}{2\sqrt{16-x^{2}}}-\left(\frac{16-2x^{2}}{2\sqrt{16-x^{2}}}\right)\)

=\(\displaystyle \L\\\frac{16-16+2x^{2}}{2\sqrt{16-x^{2}}}\)

=\(\displaystyle \H\\\fbox{\frac{x^{2}}{\sqrt{16-x^{2}}}}\)...as Soroban says, TA...DAA!.

Except, I don't believe there should be a negative as the book says.


EDIT: That Soroban beat me to it. Oh well, double confirmation.