Question
Find the derivative of the function $y=f(x)$ using the derivative of the inverse function $x=f^{-1}(y)$ in the following$y=\sqrt{1+\sqrt{x}}$

Answer

$
y=\sqrt{1+\sqrt{x}}
$
We first find the inverse of the function $y=f(x)$, i.e. $x$ in term of $y$.
$
\begin{aligned}
& y^2=1+\sqrt{x} \text { i.e. } \sqrt{x}=y^2-1, \therefore x=f^{-1}(y)=\left(y^2-1\right)^2 \\
& \begin{aligned}
\frac{d y}{d x}=\frac{1}{\frac{d x}{d y}}=\frac{1}{\frac{d}{d y}\left[\left(y^2-1\right)^2\right]} & =\frac{1}{2\left(y^2-1\right) \frac{d}{d y}\left(y^2-1\right)} \\
= & \frac{1}{2\left(y^2-1\right)(2 y)}=\frac{1}{4 \sqrt{1+\sqrt{x}}\left[(\sqrt{1+\sqrt{x}})^2-1\right]} \\
= & \frac{1}{4 \sqrt{1+\sqrt{x}}(1+\sqrt{x}-1)}=\frac{1}{4 \sqrt{x} \sqrt{1+\sqrt{x}}}
\end{aligned}
\end{aligned}
$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free