# Math Help - Differentiation Question

1. ## Differentiation Question

Hi

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

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

After going through the process this is what i get:

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

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

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

P.S

2. Hello, Paymemoney!

1) Find $y'\!:\quad x^4 + x^3y + \frac{1}{x^4y} - 13 \;=\; 0$

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

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

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

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

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

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

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

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

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

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