Search any question & find its solution
Question:
Answered & Verified by Expert
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 $x=1$
Options:
Solution:
1517 Upvotes
Verified Answer
The correct answer is:
3
$$
\text { } \begin{aligned}
f^{\prime}(x) & =\sqrt{2 x^2-1} \\
y & =f\left(x^3\right)
\end{aligned}
$$
Differentiation w.r.t. $x$
$$
\begin{array}{rlr}
\frac{d y}{d x} & =f^{\prime}\left(x^3\right) \cdot 3 x^2 \quad \text { [Using chain rule] } \\
\left.\therefore \quad \frac{d y}{d x}\right|_{x=1} & =f^{\prime}\left(1^3\right) 3(1)^2 & \\
& =f^{\prime}(1) \cdot 3 \\
& =\sqrt{2(1)^2-1} \cdot 3 \quad\left[\because f^{\prime}(x)=\sqrt{2 x^2-1}\right] \\
& =\sqrt{2-1} \cdot 3=\sqrt{1} \cdot 3 \\
& =3
\end{array}
$$
\text { } \begin{aligned}
f^{\prime}(x) & =\sqrt{2 x^2-1} \\
y & =f\left(x^3\right)
\end{aligned}
$$
Differentiation w.r.t. $x$
$$
\begin{array}{rlr}
\frac{d y}{d x} & =f^{\prime}\left(x^3\right) \cdot 3 x^2 \quad \text { [Using chain rule] } \\
\left.\therefore \quad \frac{d y}{d x}\right|_{x=1} & =f^{\prime}\left(1^3\right) 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}\right] \\
& =\sqrt{2-1} \cdot 3=\sqrt{1} \cdot 3 \\
& =3
\end{array}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.