PUC Science 2nd PUC Class 12 - Karnataka Board PUC Question Bank Solutions for Mathematics

If \[y = \cos^{- 1} \left( 2x \right) + 2 \cos^{- 1} \sqrt{1 - 4 x^2}, - \frac{1}{2} < x < 0, \text{ find } \frac{dy}{dx} \] ?

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

If \[y = \cos^{- 1} \left\{ \frac{2x - 3 \sqrt{1 - x^2}}{\sqrt{13}} \right\}, \text{ find } \frac{dy}{dx}\] ?

Differentiate \[\sin^{- 1} \left\{ \frac{2^{x + 1} \cdot 3^x}{1 + \left(36 \right)^x} \right\}\] with respect to x.

If \[y = \sin^{- 1} \left( 6x\sqrt{1 - 9 x^2} \right), - \frac{1}{3\sqrt{2}} < x < \frac{1}{3\sqrt{2}}\] \[\frac{dy}{dx} \] ?

Find \[\frac{dy}{dx}\] in the following case \[xy = c^2\]  ?

Find  \[\frac{dy}{dx}\] in the following case: \[y^3 - 3x y^2 = x^3 + 3 x^2 y\] ?

Find  \[\frac{dy}{dx}\] in the following case  \[x^{2/3} + y^{2/3} = a^{2/3}\] ?

Find  \[\frac{dy}{dx}\] in the following case \[4x + 3y = \log \left( 4x - 3y \right)\] ?

Find  \[\frac{dy}{dx}\] in the following case \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\] ?

Find  \[\frac{dy}{dx}\] in the following case \[x^5 + y^5 = 5 xy\] ?

Find  \[\frac{dy}{dx}\] in the following case \[\left( x + y \right)^2 = 2axy\] ?

Find  \[\frac{dy}{dx}\] in the following case \[\left( x^2 + y^2 \right)^2 = xy\] ?

Find  \[\frac{dy}{dx}\] in the following case \[\tan^{- 1} \left( x^2 + y^2 \right) = a\] ?

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 \[2\frac{dy}{dx} + y^3 = 0\] ?