Question

How do you find the derivative of:
(x^3 -3x^2 + 4) / x^2

Answer 1

\[ (x^3 -3x^2 + 4) \div x^2\]

Answer 2

I get, 1-8x^-3... so 1/-8x^3

Answer is x^3 - 8 / x^3

Answer 3

You just split them
\[\frac{ x^3 -3x^2 + 4 }{ x^2 } \\ \\ \\ \\ \frac{x^3}{x^2}-\frac{3x^2}{x^2}+\frac{4}{x^2} \\ \\ x-3+\frac{4}{x^2}\]
Then use power rule

Answer 4

\[\frac{d}{dx}x^n=nx^{n-1}\]

Answer 5

That's exactly what I did..
\[1 - 8x^-3\]

Answer 6

Your problem is on differentiating
\[\frac{4}{x^2}\]

Answer 7

that becomes (4) (-2x^-3)

Answer 8

Careful, put it like this first
\[\frac{4}{x^2} \\ \\ =4x^{-2}\]
Then use power rule

Answer 9

You get \(-8x^{-3}\)

Answer 10

Yes

Answer 11

Putting it all together
\(1-8x^{-3}\)

Answer 12

Yeeah, thats what I got 1-8x^-3

Answer 13

Nvm... Got it.. 1 becomes x^3 / x^3