Correct choice is (b) \(log2x^{sin3x} \,(3 \,cos3x \,log(log2x)+\frac{sin3x}{x \,log2x})\)
The explanation is: Consider y=\((log2x)^{sin3x}\)
Applying log on both sides, we get
logy=\(log(log2x)^{sin3x}\)
logy=sin3x log(log2x)
Differentiating with respect to x, we get
\(\frac{1}{y} \,\frac{dy}{dx}=log(log2x)\frac{d}{dx} (sin3x)+sin3x \frac{d}{dx} \,(log(log2x))\)
By using chain rule, we get
\(\frac{1}{y} \,\frac{dy}{dx}=log(log2x).3 \,cos3x+sin3x.\frac{1}{log2x}.\frac{1}{2x}.2 \,(∵u.v=u’ \,v+uv’)\)
\(\frac{dy}{dx}\)=y(3 cos3x log(log2x)+\(\frac{sin3x}{x \,log2x}\))
∴\(\frac{dy}{dx}\)=log2x^sin3x \(\left (3 \,cos3x \,log(log2x)+\frac{sin3x}{x \,log2x} \right )\)