Recent questions tagged exercise5-3

Solve the equation : $\;x^{2}+\large\frac{x}{\sqrt{2}}+1=0\;$

asked Apr 11, 2014 by yamini.v

1 answer

Solve the equation : $\;x^{2}+x+\large\frac{1}{\sqrt{2}}=0\;$

asked Apr 11, 2014 by yamini.v

1 answer

Solve the equation : $\;\sqrt{3}x^{2}-\sqrt{2}x+3\sqrt{3}=0\;$

asked Apr 11, 2014 by yamini.v

1 answer

Solve the equation : $\;\sqrt{2}x^{2}+x+\sqrt{2}=0\;$

asked Apr 11, 2014 by yamini.v

1 answer

Solve the equation : $\;x^{2}-x+2=0\;$

asked Apr 11, 2014 by yamini.v

1 answer

Solve the equation : $\;x^{2}+3x+5=0\;$

asked Apr 11, 2014 by yamini.v

1 answer

Solve the equation : $\;-x^{2}+x-2=0\;$

asked Apr 11, 2014 by yamini.v

1 answer

Solve the equation : $\;x^{2}+3x+9=0\;$

asked Apr 11, 2014 by yamini.v

1 answer

Solve the equation : $\;2x^{2}+x+1=0\;$

asked Apr 10, 2014 by yamini.v

1 answer

Solve the equation : $\;x^{2}+3=0\;$

asked Apr 10, 2014 by yamini.v

1 answer

Using Rolle's theorem find the points on the curve $y=x^{2}+1,-2\leq x \leq 2$ where the tangent is parallel to $x$- axis.

asked May 3, 2013 by poojasapani_1

1 answer

Verify Rolle's theorem for the following function; $f(x)=4x^{3}-9x;-\large\frac{3}{2}\leq x\leq \frac{3}{2}$.

asked May 3, 2013 by poojasapani_1

1 answer

Verify Rolle's theorem for the following functions; $f(x)=|x-1|,0\leq x\leq 2.$

asked May 3, 2013 by poojasapani_1

1 answer

Verify Rolle's theorem for the following function; $f(x)=x^{2}, 0\leq x\leq 1$

asked May 3, 2013 by poojasapani_1

1 answer

Verify Rolle's theorem for the following function; $f(x)=\sin x, 0\leq x\leq\pi$.

asked May 3, 2013 by poojasapani_1

1 answer

Find $\large \frac {dy}{dx} $ in the following: $ y = sec^{-1} \left(\large\frac{1}{2x^2 - 1}\right), 0 < x < \; { \frac{1}{\sqrt2} } \\ $

asked Nov 16, 2012 by thanvigandhi_1

1 answer

Find $\large \frac {dy}{dx} $ in the following: $y = \sin^{-1} \big( 2x \sqrt{1 -x^2} \big), -\;{ \frac{1}{\sqrt2} }\;< x < \;{\frac{1}{\sqrt2} }\\ $

asked Nov 16, 2012 by thanvigandhi_1

1 answer

Find $ \large\frac {dy}{dx} $ in the following: $ y = cos^{-1} \left(\large\frac{2x}{1 + x^2}\right), -1 < x < 1 $

asked Nov 15, 2012 by thanvigandhi_1

1 answer

Find $ \frac {dy}{dx} $ in the following:$ y = sin^{-1} \left(\frac{1 - x^2}{1 + x^2}\right), 0 < x < 1 $

asked Nov 15, 2012 by thanvigandhi_1

1 answer

Find $ \frac {dy}{dx} $ in the following: $y = cos^{-1} \left(\frac{1 - x^2}{1 + x^2}\right), 0 < x < 1 $

asked Nov 15, 2012 by thanvigandhi_1

1 answer

Find $ \frac {dy}{dx}$ in the following: $y = tan^{-1} \left(\frac{3x - x^3}{1 - 3x^2}\right), - \; { \frac{1}{\sqrt3} } < x < \;{\frac{1}{\sqrt3} }\\ $

asked Nov 15, 2012 by thanvigandhi_1

1 answer

Find $ \frac {dy}{dx} $ in the following: $ y = sin^{-1} \left(\frac{2x}{1 + x^2}\right) $

asked Nov 15, 2012 by thanvigandhi_1

1 answer

Find $ \frac {dy}{dx} $ in the following: $ \sin^2x +\cos^2y = 1 $

asked Nov 11, 2012 by thanvigandhi_1

1 answer

Find $ \frac {dy}{dx} $ in the following: $ sin^2y + cos \: xy = k $

asked Nov 11, 2012 by thanvigandhi_1

1 answer

Find $ \frac {dy}{dx} $ in the following: $ x^2 +xy + y^2 = 100 $

asked Nov 11, 2012 by thanvigandhi_1

1 answer

Find $ \frac {dy}{dx} $ in the following: $ xy +y^2 = tan \: x + y $

asked Nov 11, 2012 by thanvigandhi_1

1 answer

Find $\frac {dy}{dx}$ in the following: $ax +by^2 = cos \: y$

asked Nov 11, 2012 by thanvigandhi_1

1 answer

Find $\frac {dy}{dx}$ in the following: $2x +3y = sin \: y$

asked Nov 11, 2012 by thanvigandhi_1

1 answer

Find $\frac {dy}{dx}$ in the following: $2x +3y = sin \: x$

asked Nov 11, 2012 by thanvigandhi_1

1 answer

Ask Questions, Get Answers

Solve the equation : $\;x^{2}+\large\frac{x}{\sqrt{2}}+1=0\;$

Solve the equation : $\;x^{2}+x+\large\frac{1}{\sqrt{2}}=0\;$

Solve the equation : $\;\sqrt{3}x^{2}-\sqrt{2}x+3\sqrt{3}=0\;$

Solve the equation : $\;\sqrt{2}x^{2}+x+\sqrt{2}=0\;$

Solve the equation : $\;x^{2}-x+2=0\;$

Solve the equation : $\;x^{2}+3x+5=0\;$

Solve the equation : $\;-x^{2}+x-2=0\;$

Solve the equation : $\;x^{2}+3x+9=0\;$

Solve the equation : $\;2x^{2}+x+1=0\;$

Solve the equation : $\;x^{2}+3=0\;$

Using Rolle's theorem find the points on the curve $y=x^{2}+1,-2\leq x \leq 2$ where the tangent is parallel to $x$- axis.

Verify Rolle's theorem for the following function; $f(x)=4x^{3}-9x;-\large\frac{3}{2}\leq x\leq \frac{3}{2}$.

Verify Rolle's theorem for the following functions; $f(x)=|x-1|,0\leq x\leq 2.$

Verify Rolle's theorem for the following function; $f(x)=x^{2}, 0\leq x\leq 1$

Verify Rolle's theorem for the following function; $f(x)=\sin x, 0\leq x\leq\pi$.

Find $\large \frac {dy}{dx} $ in the following: $ y = sec^{-1} \left(\large\frac{1}{2x^2 - 1}\right), 0 < x < \; { \frac{1}{\sqrt2} } \\ $

Find $\large \frac {dy}{dx} $ in the following: $y = \sin^{-1} \big( 2x \sqrt{1 -x^2} \big), -\;{ \frac{1}{\sqrt2} }\;< x < \;{\frac{1}{\sqrt2} }\\ $

Find \( \large\frac {dy}{dx} \) in the following: $ y = cos^{-1} \left(\large\frac{2x}{1 + x^2}\right), -1 < x < 1 $

Find $ \frac {dy}{dx} $ in the following:$ y = sin^{-1} \left(\frac{1 - x^2}{1 + x^2}\right), 0 < x < 1 $

Find \( \frac {dy}{dx} \) in the following: $y = cos^{-1} \left(\frac{1 - x^2}{1 + x^2}\right), 0 < x < 1 $

Find $ \frac {dy}{dx}$ in the following: $y = tan^{-1} \left(\frac{3x - x^3}{1 - 3x^2}\right), - \; { \frac{1}{\sqrt3} } < x < \;{\frac{1}{\sqrt3} }\\ $

Find $ \frac {dy}{dx} $ in the following: $ y = sin^{-1} \left(\frac{2x}{1 + x^2}\right) $

Find $ \frac {dy}{dx} $ in the following: $ \sin^2x +\cos^2y = 1 $

Find \( \frac {dy}{dx} \) in the following: \( sin^2y + cos \: xy = k \)

Find $ \frac {dy}{dx} $ in the following: $ x^2 +xy + y^2 = 100 $

Find $ \frac {dy}{dx} $ in the following: $ xy +y^2 = tan \: x + y $

Find $\frac {dy}{dx}$ in the following: $ax +by^2 = cos \: y$

Find $\frac {dy}{dx}$ in the following: $2x +3y = sin \: y$

Find $\frac {dy}{dx}$ in the following: $2x +3y = sin \: x$