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

Concept: For composite functions involving square roots, exponentials and logarithms, use the chain rule carefully. If: \[ f(x)=\sqrt{g(x)} \] then: \[ f'(x)=\frac{g'(x)}{2\sqrt{g(x)}} \]

Step 1: Define the inner function.
Let \[ g(x)=2^{2x}\log(3x-2) \] Thus, \[ f(x)=\sqrt{g(x)} \] Hence, \[ f'(x)=\frac{g'(x)}{2\sqrt{g(x)}} \]

Step 2: Differentiate \(g(x)\).
Using product rule: \[ g'(x) = \frac{d}{dx}(2^{2x})\log(3x-2) + 2^{2x}\frac{d}{dx}\log(3x-2) \] Now, \[ \frac{d}{dx}(2^{2x}) = 2^{2x}(2\log2) \] and \[ \frac{d}{dx}\log(3x-2) = \frac3{3x-2} \] Therefore, \[ g'(x) = 2^{2x}(2\log2)\log(3x-2) + 2^{2x}\frac3{3x-2} \]

Step 3: Evaluate at \(x=2\).
First compute: \[ 2^{2(2)}=2^4=16 \] and \[ \log(3(2)-2)=\log4 \] Thus, \[ g(2)=16\log4 \] Also, \[ g'(2) = 16(2\log2)(\log4) + 16\cdot\frac34 \] \[ = 32\log2\log4+12 \] Now, \[ f'(2) = \frac{32\log2\log4+12} {2\sqrt{16\log4}} \] \[ = \frac{32\log2\log4+12} {8\sqrt{\log4}} \] \[ = \frac{8\log2\log4+3} {2\sqrt{\log4}} \] Hence, \[ \boxed{ \frac{8\log2\log4+3} {2\sqrt{\log4}} } \]