Question

Find the derivative dy/dx for y=(x^2-2x/x^3+3)

Answer 1

\(\large\color{black}{ \displaystyle y=\frac{x^2-2x}{x^3+3} }\)

Answer 2

Use,

\(\large\color{black}{ \displaystyle \frac{d}{dx} \left[\frac{f(x)}{g(x)}\right] =\frac{f'(x)g(x)-f(x)g'(x)}{\left\{g(x)\right\}^2}}\)

Answer 3

\(\large\color{black}{ \displaystyle \frac{d}{dx} \left[\frac{x^2-2x}{x^3+3}\right] =\frac{(x^2-2x)'\cdot (x^3+3)-(x^2-2x) \cdot (x^3+3)'}{\left\{x^3+3\right\}^2}}\)

Answer 4

The derivative of:

[1] \(\color{black}{x^2-2x}\)
[2] \(\color{black}{x^3+3}\)

are?

Answer 5

2x-2 and 3x^2

Answer 6

Yes, so:

\(\large\color{black}{ \displaystyle \frac{d}{dx} \left[\frac{x^2-2x}{x^3+3}\right] =\frac{(2x-2)\cdot (x^3+3)-(x^2-2x) \cdot (3x^2)}{\left\{x^3+3\right\}^2}}\)

Answer 7

Now, simplify if you can...

Answer 8

\(\large\color{black}{ \displaystyle \frac{d}{dx} \left[\frac{x^2-2x}{x^3+3}\right] =\frac{(2x^4-2x^3+6x-6)-(3x^4-6x^3)}{\left\{x^3+3\right\}^2}}\)

\(\large\color{black}{ \displaystyle \frac{d}{dx} \left[\frac{x^2-2x}{x^3+3}\right] =\frac{2x^4-2x^3+6x-6-3x^4+6x^3}{\left\{x^3+3\right\}^2}}\)

\(\large\color{black}{ \displaystyle \frac{d}{dx} \left[\frac{x^2-2x}{x^3+3}\right] =\frac{-x^4+4x^3+6x-6}{x^6+6x^3+9}}\)

Answer 9

that is the most you can "simplify"

Answer 10

\(\large\color{black}{ \displaystyle \frac{dy}{dx} =\frac{-x^4+4x^3+6x-6}{x^6+6x^3+9}}\)

Answer 11

that's wwhat I got

Answer 12

so b right