If \( y=\sqrt{12^{x}(3x^{2}+2)} \), then \( \displaystyle\frac{dy}{dx} \) will be
(consider \( \log_{e}x=\log x \))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{12^{x}\left((3x^{2}+2)\log 12+6x\right)}{2y} \)
Squaring both sides removes the root:
\( \displaystyle y^{2}=12^{x}(3x^{2}+2) \)
Differentiating both sides with respect to x, using \( \frac{d}{dx}\left(12^{x}\right)=12^{x}\log 12 \) and the product rule:
\( \displaystyle 2y\frac{dy}{dx}=12^{x}\log 12\,(3x^{2}+2)+12^{x}(6x) \)
\( \displaystyle \frac{dy}{dx}=\frac{12^{x}\left((3x^{2}+2)\log 12+6x\right)}{2y} \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{12^{x}\left((3x^{2}+2)\log 12+6x\right)}{y} \) — Incorrect. The factor 2 from differentiating \( y^{2} \) (that is, 2y dy/dx) has been dropped from the denominator.
Option (3) \( \displaystyle\frac{12^{x-1}\left(x(3x^{2}+2)+6x\right)}{2y} \) — Incorrect. \( 12^{x} \) has been differentiated as if it were a power of x, giving \( x\,12^{x-1} \); the derivative of \( a^{x} \) is \( a^{x}\log a \). The term 6x is also multiplied by \( 12^{x-1} \) instead of \( 12^{x} \).
Option (4) \( \displaystyle\frac{x12^{x-1}(3x^{2}+2)+6x\,12^{x}}{2y} \) — Incorrect. Here too \( 12^{x} \) has been differentiated as \( x\,12^{x-1} \) instead of \( 12^{x}\log 12 \).