Question

If \[ f(x)=\sqrt{2^{2x}\log(3x-2)} \] then \(f'(2)=\)

• A. \(\dfrac{4\log2\log4+3}{\sqrt{\log4}}\)
• B. \(\dfrac{8(\log2)^2+3}{2\sqrt{\log4}}\)
• C. \(\dfrac{8(\log4)^2+3}{2\sqrt{\log4}}\)
• D. \(\dfrac{8\log2\log4+3}{2\sqrt{\log4}}\)

Hint:

For expressions involving: \[ \sqrt{g(x)} \] always use: \[ \frac{d}{dx}\sqrt{g(x)} = \frac{g'(x)}{2\sqrt{g(x)}} \] Also remember: \[ \frac{d}{dx}(a^x)=a^x\log a \]

Answer 1

The Correct Option is D

Step 1: Set up the chain rule.
Let $g(x)=2^{2x}\log(3x-2)$ so that $f(x)=\sqrt{g(x)}$. Then $f'(x)=\dfrac{g'(x)}{2\sqrt{g(x)}}$.
Step 2: Differentiate the exponential part.
$\dfrac{d}{dx}\,2^{2x}=2^{2x}\cdot 2\log2$.
Step 3: Differentiate the log part.
$\dfrac{d}{dx}\log(3x-2)=\dfrac{3}{3x-2}$.
Step 4: Apply the product rule for $g'(x)$.
\[ g'(x)=2^{2x}(2\log2)\log(3x-2)+2^{2x}\cdot\frac{3}{3x-2}. \]
Step 5: Plug in $x=2$.
Here $2^{2x}=2^4=16$ and $\log(3\cdot2-2)=\log4$. So $g(2)=16\log4$ and \[ g'(2)=16(2\log2)(\log4)+16\cdot\frac34=32\log2\log4+12. \]
Step 6: Build $f'(2)$.
\[ f'(2)=\frac{32\log2\log4+12}{2\sqrt{16\log4}}=\frac{32\log2\log4+12}{8\sqrt{\log4}}=\frac{8\log2\log4+3}{2\sqrt{\log4}}. \] \[ \boxed{\frac{8\log2\log4+3}{2\sqrt{\log4}}} \]