Question

If $f\text{'}\left(x\right)=\sqrt{2x^2-1}\mathrm{and}y=f\left(x^2\right)$, then find $\frac{dy}{dx}atx=1$.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},f\text{'}\left(x\right)=\sqrt{2x^2-1} \\ \mathrm{and}y=f\left(x^2\right) \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{dy}{dx}=\frac{d}{dx}f\left(x^2\right) \\ \Rightarrow \frac{dy}{dx}=f\text{'}\left(x^2\right)\frac{d}{dx}\left(x^2\right) \\ \Rightarrow \frac{dy}{dx}=f\text{'}\left(x^2\right)\times 2x \\ \Rightarrow \frac{dy}{dx}=2xf\text{'}\left(x^2\right) \\ \mathrm{Putting}x=1,\mathrm{we}\mathrm{get}, \\ \frac{dy}{dx}=2\left(1\right)f\text{'}\left(1^2\right) \\ \Rightarrow \frac{dy}{dx}=2\times f\text{'}\left(1\right) \\ \Rightarrow \frac{dy}{dx}=2\times 1\left[\because f\text{'}\left(1\right)=\sqrt{2{\left(1\right)}^2-1}=\sqrt{2-1}=1\right] \\ \Rightarrow \frac{dy}{dx}=2 \end{gathered}$$