If tan inverse y/x = log root x^2 + y^2, then prove that

Question:

If $\tan^{-1}(\frac{y}{x}) = \log \sqrt{x^2+y^2}$ , Prove that $\dfrac{dy}{dx} = \dfrac{x + y}{x - y}$ . ……..[CBSC 2020]

Solution:

Given, $\tan^{-1}(\frac{y}{x}) = \log \sqrt{x^2+y^2}$

$\Rightarrow \tan^{-1}(\frac{y}{x}) = \frac{1}{2}\log (x^2 + y^2)$

Differentiating with respect to x, we have,

$\Rightarrow \dfrac{1 }{1+ (\frac{y}{x})^2}\times \left[\dfrac{x(\frac{y}{x})-1\times 1}{x^2}\right] = \dfrac{1}{2}\times \dfrac{1}{x^2+y^2}\times (2x + 2y(\frac{dy}{dx}))$

$\Rightarrow \dfrac{x^2}{x^2+y^2}\times \dfrac{\left(x\frac{dy}{dx}-y\right)}{x^2} = \dfrac{1}{x^2 + y^2}\left(x + y\frac{dy}{dx}\right)$

$\Rightarrow x\dfrac{dy}{dx} - y = x + y\dfrac{dy}{dx}$

$\Rightarrow (x - y)\dfrac{dy}{dx} = x + y$

$\Rightarrow \dfrac{dy}{dx} = \dfrac{x + y}{x - y}$

Some other question:

Q 1: If $y^x = e^{(y-x)}$ , then prove that $dy/dx = (1 + \log y)^2\log y$

Q 2: Find the value of $\dfrac{dy}{dx}$ at $\theta = \dfrac{\pi}{4}$ , if $x = ae^{\theta}(\sin \theta - \cos \theta)$ and $y = ae^{\theta}(\sin \theta + \cos \theta)$ . …….. [CBSC 2008, 2014]

If tan inverse y/x = log root x^2 + y^2, then prove that

Question:

If $\tan^{-1}(\frac{y}{x}) = \log \sqrt{x^2+y^2}$ , Prove that $\dfrac{dy}{dx} = \dfrac{x + y}{x - y}$ . ……..[CBSC 2020]

Solution:

Some other question:

Team Gmath

(1 + e^{y/x})dy + e^{y/x}(1 – \frac{y}{x}) = 0

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A child puts one five-rupees coin of her savings in the

Yasmeen saves ₹ 32 during the first month ₹ 36

Show that the sum of all terms of an A.P. whose first term is a

If the sum of first ‘p’ terms of an A.P. is ‘q’

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