Solution Q. 10

 

Given the equation $x=e^{{\tan }^{-1}\left(\frac{y-x^2}{x^2}\right)}$, 

we need to find $\left(\frac{dy}{dx}\right)$ at $x=1$.

First, we take the natural logarithm of both sides:

$$\ln x={\tan }^{-1}\left(\frac{y-x^2}{x^2}\right)$$

Let $u=\frac{y-x^2}{x^2}$. Then the equation becomes:

$$\ln x={\tan }^{-1}(u)$$

Taking the tangent of both sides, we get:

$$u=\tan (\ln x)$$

Substituting back $u=\frac{y-x^2}{x^2}$, we have:

$$\frac{y-x^2}{x^2}=\tan (\ln x)$$

Solving for $y$:

$$y=x^2(1+\tan (\ln x))$$

Next, we differentiate $y$ with respect to $x$:

$$\frac{dy}{dx}=\frac{d}{dx}[x^2(1+\tan (\ln x))]$$

Using the product rule, we get:

$$\frac{dy}{dx}=2x(1+\tan (\ln x))+x^2({\sec }^2(\ln x)\cdot \frac{1}{x})$$

Simplifying the second term:

$$\frac{dy}{dx}=2x(1+\tan (\ln x))+x{\sec }^2(\ln x)$$

Evaluating at $x=1$:

• $\ln (1)=0$

• $\tan (0)=0$

• ${\sec }^2(0)=1$

Substituting these values:

$$\frac{dy}{dx}{\mid }_{x=1}=2\cdot 1\cdot (1+0)+1\cdot 1=2+1=3$$

Thus, the value of $\left(\frac{dy}{dx}\right)$ at $x=1$ is $\boxed{{\displaystyle 3}}$.

SImilarly find for other values of x.

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