If $g$ is the inverse of $f$ and $f^{\prime}(x)=\frac{1}{1+x^3}$, then $g^{\prime}(x)$ is

If $g$ is the inverse of $f$ and $f^{\prime}(x)=\frac{1}{1+x^3}$, then $g^{\prime}(x)$ is
  1. $\frac{1}{1+(\operatorname{g}(x))^3}$
  2. $1+(g(x))^3$
  3. $\frac{g(x)}{1+(g(x))^3}$
  4. $\frac{(g(x))^3}{1+(g(x))^3}$

Solution

$\begin{array}{ll} & \mathrm{g}(x) \text { is inverse of function } \mathrm{f}(x) \\ & \text { i.e., } \mathrm{g}(x)=\mathrm{f}^{-1}(x) \\ & \mathrm{f}(\mathrm{g}(x))=x \\ & \text { differentiating w.r.t. } x \text {, we get } \\ & \mathrm{f}^{\prime}(\mathrm{g}(x)) \times \mathrm{g}^{\prime}(x)=1 \\ \therefore \quad & \mathrm{g}^{\prime}(x)=\frac{1}{\mathrm{f}^{\prime}(\mathrm{g}(x))}... (i)\end{array}$ Now, $\mathrm{f}^{\prime}(x)=\frac{1}{1+x^3}$... [Given] $\therefore \quad \mathrm{f}^{\prime}(\mathrm{g}(x))=\frac{1}{1+(\mathrm{g}(x))^3}$ ... (ii) $\therefore \quad$ From (i) and (ii), we get $\begin{aligned} & \therefore \quad \mathrm{g}^{\prime}(x)=1+(\mathrm{g}(x))^3 \\ & \end{aligned}$

Asked in: MHT CET 2023 (09 May Shift 2)

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