WAEC - Further Mathematics (2022 - No. 12)
Differentiate \(\frac{5x^ 3+x^2}{x}\), x ≠ 0 with respect to x.
10x + 1
10x + 2
x(15x + 1)
x(15x + 2)
คำอธิบาย
\(\frac{5x^ 3+x^2}{x}\) → \(\frac{5x^ 3}{x} + \frac{x^2}{x}\)
5x\(^2\) + x
Then dy/dx = 10x + 1
\(\frac{5x^ 3+x^2}{x}\) → \(\frac{5x^ 3}{x} + \frac{x^2}{x}\)
5x\(^2\) + x
Then dy/dx = 10x + 1