CA Foundation · Quantitative Aptitude · Differential and Integral Calculus
If y = e^(3x) + log x, then d²y/dx² is:
The second derivative is 9e^(3x) − 1/x². The first derivative is 3e^(3x) + 1/x. Differentiating again, the exponential term picks up another factor of 3 giving 9e^(3x), while 1/x differentiates to −1/x².
- A3e^(3x) − 1/x²
- B9e^(3x) + 1/x²
- C9e^(3x) − 1/x²Correct
- D3e^(3x) + 1/x²
Explanation
dy/dx = 3e^(3x) + 1/x. Differentiating again: d/dx of 3e^(3x) = 9e^(3x), and d/dx of x⁻¹ = −x⁻². So d²y/dx² = 9e^(3x) − 1/x². Option 2 has the wrong sign for the log term.
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