# Differentiation Question

#### Paymemoney

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.

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

P.S

#### Soroban

MHF Hall of Honor
Hello, Paymemoney!

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

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 . . .

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

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

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