If $f^{\prime}(x)=\sqrt{2 x^2-1}$ and $y=f\left(x^3\right)$, then find the value of $\frac{d y}{d x}$ at $x=1$

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Aicharya · Accepted Answer

$ 	ext { } \begin{aligned} f^{\prime}(x) & =\sqrt{2 x^2-1} \ y & =f\left(x^3ight) \end{aligned} $ Differentiation w.r.t. $x$ $ \begin{array}{rlr} \frac{d y}{d x} & =f^{\prime}\left(x^3ight) \cdot 3 x^2 \quad 	ext { [Using chain rule] } \ \left.	herefore \quad \frac{d y}{d x}ight|_{x=1} & =f^{\prime}\left(1^3ight) 3(1)^2 & \ & =f^{\prime}(1) \cdot 3 \ & =\sqrt{2(1)^2-1} \cdot 3 \quad\left[\because f^{\prime}(x)=\sqrt{2 x^2-1}ight] \ & =\sqrt{2-1} \cdot 3=\sqrt{1} \cdot 3 \ & =3 \end{array} $

If $f^{\prime}(x)=\sqrt{2 x^2-1}$ and $y=f\left(x^3\right)$, then find the value of $\frac{d y}{d x}$ at $x=1$

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