JAMB - Mathematics (1997 - No. 32)
Differentiate \(\frac{6x^3 - 5x^2 + 1}{3x^2}\) with respect to x
\(\frac{2 + 2}{3x^3}\)
2 + \(\frac{1}{6x}\)
2 - \(\frac{2}{3x^3}\)
\(\frac{1}{5}\)
ব্যাখ্যা
\(\frac{6x^3 - 5x^2 + 1}{3x^2}\)
let y = 3x2
y = \(\frac{6x^3}{3x^2}\) - \(\frac{6x^2}{3x^2}\) + \(\frac{1}{3x^2}\)
Y = 2x - \(\frac{5}{3}\) + \(\frac{1}{3x^2}\)
\(\frac{dy}{dx}\) = 2 + \(\frac{1}{3}\)(-2)x-3
= 2 - \(\frac{2}{3x^3}\)
let y = 3x2
y = \(\frac{6x^3}{3x^2}\) - \(\frac{6x^2}{3x^2}\) + \(\frac{1}{3x^2}\)
Y = 2x - \(\frac{5}{3}\) + \(\frac{1}{3x^2}\)
\(\frac{dy}{dx}\) = 2 + \(\frac{1}{3}\)(-2)x-3
= 2 - \(\frac{2}{3x^3}\)