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If f(x) = sin⁻¹((2x 3ˣ)/(1 + 9ˣ)), then f′(-1/2) equals -

1. √3 loge √3
2. -√3 loge √3
3. -√3 loge 3
4. √3 loge 3

Correct answer: √3 loge √3

Solution

With t=3^x, f(x)=sin^-1(2*3^x/(1+9^x))=2 arctan(3^x), so f'(x)=2*3^x ln3/(1+9^x). At x=-1/2: 3^x=1/sqrt(3), 9^x=1/3, giving f'=(sqrt(3)/2)ln3 = sqrt(3) loge(sqrt(3)).

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