Arts (English Medium) Class 12 - CBSE Question Bank Solutions for Mathematics

Find  \[\frac{dy}{dx}\] in the following case \[e^{x - y} = \log \left( \frac{x}{y} \right)\] ?

Find  \[\frac{dy}{dx}\] in the following case \[\sin xy + \cos \left( x + y \right) = 1\] ?

If \[\sqrt{1 - x^2} + \sqrt{1 - y^2} = a \left( x - y \right)\] , prove that \[\frac{dy}{dx} = \frac{\sqrt{1 - y^2}}{1 - x^2}\] ?

If `ysqrt(1-x^2) + xsqrt(1-y^2) = 1` prove that `dy/dx = -sqrt((1-y^2)/(1-x^2))`

If \[xy = 1\] prove that \[\frac{dy}{dx} + y^2 = 0\] ?

If \[x y^2 = 1,\] prove that \[2\frac{dy}{dx} + y^3 = 0\] ?

If \[x \sqrt{1 + y} + y \sqrt{1 + x} = 0\] , prove that \[\left( 1 + x \right)^2 \frac{dy}{dx} + 1 = 0\]  ?

If \[\log \sqrt{x^2 + y^2} = \tan^{- 1} \left( \frac{y}{x} \right)\] Prove that \[\frac{dy}{dx} = \frac{x + y}{x - y}\] ?

If \[\sec \left( \frac{x + y}{x - y} \right) = a\] Prove that  \[\frac{dy}{dx} = \frac{y}{x}\] ?

If \[\tan^{- 1} \left( \frac{x^2 - y^2}{x^2 + y^2} \right) = a\] Prove that  \[\frac{dy}{dx} = \frac{x}{y}\frac{\left( 1 - \tan a \right)}{\left( 1 + \tan a \right)}\] ?

If \[xy \log \left( x + y \right) = 1\] ,Prove that \[\frac{dy}{dx} = - \frac{y \left( x^2 y + x + y \right)}{x \left( x y^2 + x + y \right)}\] ?

If \[y = x \sin \left( a + y \right)\] ,Prove that \[\frac{dy}{dx} = \frac{\sin^2 \left( a + y \right)}{\sin \left( a + y \right) - y \cos \left( a + y \right)}\] ?

If \[x \sin \left( a + y \right) + \sin a \cos \left( a + y \right) = 0\] Prove that \[\frac{dy}{dx} = \frac{\sin^2 \left( a + y \right)}{\sin a}\] ?

If \[y = x \sin y\] , Prove that \[\frac{dy}{dx} = \frac{\sin y}{\left( 1 - x \cos y \right)}\] ?

If \[y \sqrt{x^2 + 1} = \log \left( \sqrt{x^2 + 1} - x \right)\] ,Show that \[\left( x^2 + 1 \right) \frac{dy}{dx} + xy + 1 = 0\] ?

If \[\sin \left( xy \right) + \frac{y}{x} = x^2 - y^2 , \text{ find}  \frac{dy}{dx}\] ?

If  \[\tan \left( x + y \right) + \tan \left( x - y \right) = 1, \text{ find}  \frac{dy}{dx}\] ?

If \[e^x + e^y = e^{x + y} , \text{ prove that } \frac{dy}{dx} = - \frac{e^x \left( e^y - 1 \right)}{e^y \left( e^x - 1 \right)} or \frac{dy}{dx} + e^{y - x} = 0\] ?

If \[\cos y = x \cos \left( a + y \right), \text{ with } \cos a \neq \pm 1, \text{ prove that } \frac{dy}{dx} = \frac{\cos^2 \left( a + y \right)}{\sin a}\] ?

If \[\sin^2 y + \cos xy = k,\] find  \[\frac{dy}{dx}\] at \[x = 1 , \] \[y = \frac{\pi}{4} .\]