Differentiation Question

Dec 2008
509
2
Hi

I need help on the following questions. I am not sure if the answer i got is correct.

1)Find the derivative \(\displaystyle y'(x) of x^4 + y(x^3) + \frac{1}{y(x)^4} - 13 =0\)

After going through the process this is what i get:

\(\displaystyle \frac{4y^2x^3 + 3y^3x^2 + \frac{dy}{dx}x^3y^2 - 4yx^{-5} - x^{-4} \frac{dy}{dx}}{y^2}\)

2)Find the derivative of: \(\displaystyle \sqrt{\frac{x+8}{(2x+5)(x^2+2}}\)
\(\displaystyle
{\frac{x+8}{(2x+5)(x^2+2}}^{0.5}\)

Firstly i used to chain rule and then inside the square root i used the quotient rule.

Eventually i get this answer:

\(\displaystyle {\frac{2x^3-x^2-6x-6}{2\frac{x+8}{(2x+5)(x^2+2)}(2x+5)^2(x^2+2)^2}}\)

P.S
 

Soroban

MHF Hall of Honor
May 2006
12,028
6,341
Lexington, MA (USA)
Hello, Paymemoney!

1) Find \(\displaystyle y'\!:\quad x^4 + x^3y + \frac{1}{x^4y} - 13 \;=\; 0\)
Sorry, your answer makes no sense.


We have: .\(\displaystyle x^4 + x^3y + x^{-4}y^{-1} - 13 \;=\;0\)

Differentiate implicitly: .\(\displaystyle 4x^3 + x^3(y') + 3x^2y - x^{-4}y^{-2}(y') - 4x^{-5}y^{-1} \;=\;0 \)

Multiply by \(\displaystyle x^5y^2\!:\quad 4x^8y^2 + x^8y^2(y') + 3x^7y^3 - x(y') - 4y \;=\;0 \)

Rearrange terms: . \(\displaystyle x^8y^2(y') - x(y') \;=\;4y - 4x^8y^2 - 3x^7y^3\)

Factor: . \(\displaystyle x(x^7y^2-1)\cdot y' \;=\;y(4 - 4x^7y - 3x^7y^2) \)


Therefore: . \(\displaystyle y' \;=\;\frac{y(4 - 4x^8y - 3x^7y^2)}{x(x^7y^2-1)} \)




2) Diffrerentiate: . \(\displaystyle y \;=\;\sqrt{\frac{x+8}{(2x+5)(x^2+2)}} \;=\;\left(\frac{x+8}{2x^3+5x^2+4x+10}\right)^{\frac{1}{2}}\)
Your game plan is correct, but . . .


\(\displaystyle y' \;=\;\frac{1}{2}\left(\frac{x+8}{2x^3+5x^2+4x+10}\right)^{-\frac{1}{2}}\cdot\) \(\displaystyle \left[\frac{(2x^3+5x^2+4x+10)\cdot 1 - (x+8)(6x^2 + 10x + 4)}{(2x^3+5x^2+4x+10)^2}\right] \)

\(\displaystyle y' \;= \;\frac{1}{2}\cdot\frac{(2x^3+5x^2+4x+10)^{\frac{1}{2}}}{(x+8)^{\frac{1}{2}}}\cdot\left[\frac{-4x^3 - 9x^2 - 80x - 22}{(2x^3+5x^2+4x+10)^2}\right] \)

\(\displaystyle y' \;=\;-\frac{4x^3+9x^2+80x+22}{2(x+8)^{\frac{1}{2}}(2x^3+5x^2+4x+10)^{\frac{3}{2}}} \)