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

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

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

If  \[y = \log \sqrt{\frac{1 + \tan x}{1 - \tan x}}\]  prove that \[\frac{dy}{dx} = \sec 2x\] ?

If \[y = \sqrt{x} + \frac{1}{\sqrt{x}}\], prove that  \[2 x\frac{dy}{dx} = \sqrt{x} - \frac{1}{\sqrt{x}}\] ?

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

If \[y = \frac{e^x - e^{- x}}{e^x + e^{- x}}\] .prove that \[\frac{dy}{dx} = 1 - y^2\] ?

If  \[y = \left( x - 1 \right) \log \left( x - 1 \right) - \left( x + 1 \right) \log \left( x + 1 \right)\] , prove that \[\frac{dy}{dc} = \log \left( \frac{x - 1}{1 + x} \right)\] ?

If \[y = e^x \cos x\] ,prove that \[\frac{dy}{dx} = \sqrt{2} e^x \cdot \cos \left( x + \frac{\pi}{4} \right)\] ?

If \[y = \frac{1}{2} \log \left( \frac{1 - \cos 2x }{1 + \cos 2x} \right)\] , prove that \[\frac{ dy }{ dx } = 2 \text{cosec }2x \] ?

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

If \[y = \sqrt{x^2 + a^2}\] prove that  \[y\frac{dy}{dx} - x = 0\] ?

If \[y = e^x + e^{- x}\] prove that  \[\frac{dy}{dx} = \sqrt{y^2 - 4}\] ?

If \[y = \sqrt{a^2 - x^2}\] prove that  \[y\frac{dy}{dx} + x = 0\] ?

If xy = 4, prove that \[x\left( \frac{dy}{dx} + y^2 \right) = 3 y\] ?

Prove that \[\frac{d}{dx} \left\{ \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{- 1} \frac{x}{a} \right\} = \sqrt{a^2 - x^2}\] ?

Differentiate \[\cos^{- 1} \left\{ 2x\sqrt{1 - x^2} \right\}, \frac{1}{\sqrt{2}} < x < 1\] ?

Differentiate \[\cos^{- 1} \left\{ \sqrt{\frac{1 + x}{2}} \right\}, - 1 < x < 1\] ?

Differentiate  \[\sin^{- 1} \left\{ \sqrt{\frac{1 - x}{2}} \right\}, 0 < x < 1\]  ?

Differentiate \[\sin^{- 1} \left\{ \sqrt{1 - x^2} \right\}, 0 < x < 1\] ?

Differentiate \[\tan^{- 1} \left\{ \frac{x}{\sqrt{a^2 - x^2}} \right\}, - a < x < a\] ?