If f x =2+x 1+x^(1+x) , then f^(′) 0 is equal to

Question

MARKS Content Team · Accepted Answer

Correct answer: (D) 2log2−1. By taking logarithm on both sides, logf(x) = (1+x) log(2+x)−log(1+x) On differentiating, we get, 1/fx·f^(')x=(1+x)/(2+x)+log(2+x)−(1+x)/(1+x)−log(1+x) Now, put x=0 f^(')(0)=f0(1+log2)/2−1=f0log2−1/2 =2log2−1 ∵f 0 =2

If $f (x) = {(\frac{2 + x}{1 + x})}^{1 + x}$ , then $f^{'} (0)$ is equal to

Solution

More Differentiation Questions

If f x = 2+x 1+x 1+x , then f ′ 0 is equal to

Solution

More Differentiation Questions

If $f (x) = {(\frac{2 + x}{1 + x})}^{1 + x}$ , then $f^{'} (0)$ is equal to