MHT CET · Maths · Functions
If \(g\) is the inverse of \(f\) and \(f^{\prime}(x)=\frac{1}{1+x^{2}}\), then \(g^{\prime}(x)\) is equal to
- A \(1+[g(x)]^{2}\)
- B \(\frac{-1}{1+[g(x)]^{2}}\)
- C \(\frac{1}{2\left(1+x^{2}\right)}\)
- D None of these
Answer & Solution
Correct Answer
(A) \(1+[g(x)]^{2}\)
Step-by-step Solution
Detailed explanation
Given, \(g=\) inverse of \(f=f^{-1}\)
\(
\Rightarrow \quad g(x)=f^{-1}(x)
\)
\(
\Rightarrow \quad f[g(x)]=x
\)
On differentiating w.r.t. \(x\), we get
\(f^{\prime}[g(x)] \cdot g^{\prime}(x) =1 \)
\( \Rightarrow \quad g^{\prime}(x) =\frac{1}{f^{\prime}[g(x)]} \)
\( =\frac{1}{\frac{1}{1+[g(x)]^{2}}} \)
\( \Rightarrow g^{\prime}(x)=1+[g(x)]^{2}\)
\(
\Rightarrow \quad g(x)=f^{-1}(x)
\)
\(
\Rightarrow \quad f[g(x)]=x
\)
On differentiating w.r.t. \(x\), we get
\(f^{\prime}[g(x)] \cdot g^{\prime}(x) =1 \)
\( \Rightarrow \quad g^{\prime}(x) =\frac{1}{f^{\prime}[g(x)]} \)
\( =\frac{1}{\frac{1}{1+[g(x)]^{2}}} \)
\( \Rightarrow g^{\prime}(x)=1+[g(x)]^{2}\)
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