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If $\mathrm{f}$ and $\mathrm{g}$ are differentiable functions satisfying $\mathrm{g}^{\prime}(\mathrm{a})=2, \mathrm{~g}(\mathrm{a})=\mathrm{b}$ and $\mathrm{fog}=\mathrm{I}$, where is an identity function, then $\mathrm{f}^{\prime}(\mathrm{b})$ is equal to
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Verified Answer
The correct answer is:
$\frac{1}{2}$
Given
$$
g(a)=b, g^{\prime}(a)=2, f[g(x)]=x
$$
Now $f^{\prime}[g(x)] g^{\prime}(x)=1 \Rightarrow f^{\prime}(g(x))=\frac{1}{g^{\prime}(x)}$
Put $x=a$, we get
$$
f^{\prime}[g(a)]=\frac{1}{g^{\prime}(a)} \Rightarrow f^{\prime}(b)=\frac{1}{2}
$$
$$
g(a)=b, g^{\prime}(a)=2, f[g(x)]=x
$$
Now $f^{\prime}[g(x)] g^{\prime}(x)=1 \Rightarrow f^{\prime}(g(x))=\frac{1}{g^{\prime}(x)}$
Put $x=a$, we get
$$
f^{\prime}[g(a)]=\frac{1}{g^{\prime}(a)} \Rightarrow f^{\prime}(b)=\frac{1}{2}
$$
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