Question:
The derivative of a function \(f(x) = \dfrac{3x^3 + \log x}{x^2} + \tan x\) is
फलन \(f(x) = \dfrac{3x^3 + \log x}{x^2} + \tan x\) का अवकलज (derivative) क्या है?
Options:
- \(\dfrac{x^3 + 1 – \log x^2}{x^3} – \sec^2 x\)
- \(\dfrac{3x^3 – 1 – \log x^2}{x^2} + \sec^2 x\)
- \(\dfrac{x^3 – 1 – \log x^2}{x^3} + \sec^2 x\)
- \(\dfrac{3x^3 + 1 – \log x^2}{x^3} + \sec^2 x\)
- \(\dfrac{x^3 + 1 – \log x^2}{x^3} – \sec^2 x\)
- \(\dfrac{3x^3 – 1 – \log x^2}{x^2} + \sec^2 x\)
- \(\dfrac{x^3 – 1 – \log x^2}{x^3} + \sec^2 x\)
- \(\dfrac{3x^3 + 1 – \log x^2}{x^3} + \sec^2 x\)
Correct Answer: d)
Explanation:
\(f(x) = 3x + \dfrac{\log x}{x^2} + \tan x\).
\(f'(x) = 3 + \dfrac{x – 2x\log x}{x^4} + \sec^2 x = \dfrac{3x^3 + 1 – \log x^2}{x^3} + \sec^2 x\).
अवकलन करने पर: \(f'(x) = \dfrac{3x^3 + 1 – \log x^2}{x^3} + \sec^2 x\)।