TS EAMCET · Maths · Differentiation
If \(f(x)=\frac{x}{1+x}\) and \(g(x)=f(f(x))\), then \(g^{\prime}(x)\) is equal to
- A \(\frac{1}{(2 x+3)^2}\)
- B \(\frac{1}{(x+1)^2}\)
- C \(\frac{1}{x^2}\)
- D \(\frac{1}{(2 x+1)^2}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{(2 x+1)^2}\)
Step-by-step Solution
Detailed explanation
Given, \(f(x)=\frac{x}{1+x}\) and \[ \begin{array}{rlrl} \text { and } & g(x) & =f(f(x)) \\ \therefore & g(x) & =f\left(\frac{x}{x+1}\right) \\ & =\frac{\frac{x}{1+x}}{1+\frac{x}{x+1}} \\ \Rightarrow \quad & g(x) & =\frac{x}{2 x+1} \end{array} \] On differentiating both sides…
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