Application of Derivatives 4 Question 55
58. Find the coordinates of the point on the curve $y=\frac{x}{1+x^{2}}$, where the tangent to the curve has the greatest slope.
(1984, 4M)
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Answer:
Correct Answer: 58. (2)
Solution:
- Given, $\quad y=\frac{x}{1+x^{2}}$
$$ \begin{aligned} & \Rightarrow \quad \frac{d y}{d x}=\frac{\left(1+x^{2}\right) \cdot 1-x(2 x)}{\left(1+x^{2}\right)^{2}}=\frac{1-x^{2}}{\left(1+x^{2}\right)^{2}} \\ & \text { Let } \quad \frac{d y}{d x}=g(x) \quad \text { [i.e. slope of tangent] } \\ & \therefore \quad g(x)=\frac{1-x^{2}}{\left(1+x^{2}\right)^{2}} \\ & \Rightarrow g^{\prime}(x)=\frac{\left(1+x^{2}\right)^{2} \cdot(-2 x)-\left(1-x^{2}\right) \cdot 2\left(1+x^{2}\right) \cdot 2 x}{\left(1+x^{2}\right)^{4}} \\ & =\frac{-2 x\left(1+x^{2}\right)\left[\left(1+x^{2}\right)+2\left(1-x^{2}\right)\right]}{\left(1+x^{2}\right)^{4}}=\frac{-2 x\left(3-x^{2}\right)}{\left(1+x^{2}\right)^{3}} \end{aligned} $$
For greatest or least values of $m$, we should have
$$ g^{\prime}(x)=0 \Rightarrow x=0, x= \pm \sqrt{3} $$
Now,
$$ g^{\prime \prime}(x)=\frac{\left(1+x^{2}\right)^{3}\left(6 x^{2}-6\right)-\left(2 x^{3}-6 x\right) \cdot 3\left(1+x^{2}\right)^{2} \cdot 2 x}{\left(1+x^{2}\right)^{6}} $$
At $\quad x=0, g^{\prime \prime}(x)=-6<0$
$\therefore g^{\prime}(x)$ has a maximum value at $x=0$.
$\Rightarrow(x=0, y=0)$ is the required point at which tangent to the curve has the greatest slope.
