Derivative (1st order)

Study Material (PDF)
Self Assesment
Study Material (PDF)
Self Assesment

Click on each question to see answer & solution.

Question No: #121

If `y=tan^{-1}\frac{1}{1+x+x^2}+tan^{-1}\frac{1}{3+3x+x^2}+cot^{-1](7+5x+x^2)` then show that `(\frac{dy}{dx})_{x=0}=-\frac{9}{16}`

Question No: #122

If `y=tan^{-1]\frac{1}{1+x+x^2}+tan^{-1]\frac{1}{x^2+3x+3}+tan^{-1]\frac{1}{x^2+5x+7}+\ \cdots` upto `n` terms, then find the value of `\frac{dy}{dx}`

Question No: #123

If `y=tan^{-1}\frac{3x}{1+4x^2}+tan^{-1}\frac{2+5x}{5-2x}`, then show that `\frac{dy}{dx}=\frac{4}{1+16x^2}`

Question No: #124

If `y=tan^{-1}\frac{4x}{1+5x^2}+tan^{-1}\frac{2+3x}{3-2x}` then show that `\frac{dy}{dx}=\frac{5}{1+25x^2}`

Question No: #125

If `y=sin(cos^{-1}x)+\frac{1}{2}sin^{-1}\frac{2x}{1+x^2}`, then show that `\frac{dy}{dx}+\frac{1}{\sqrt{1-x^2}}=\frac{1}{1+x^2}`

Question No: #126

If `y=tan^{-1}[\frac{\sqrt{1+t^2}+\sqrt{1-t^2}}{\sqrt{1+t^2}-\sqrt{1-t^2}}]` then find the value of `\frac{dy}{dt}`

Question No: #127

If `y=tan^{-1}[\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}]` then show that `\frac{dy}{dx}=\frac{x}{\sqrt{1-x^4}}`

Question No: #128

If `y=cot^{-1}[\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}]` then find the value of `\frac{dy}{dx}`

Question No: #129

If `y=cot^{-1}[\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}]`, then show that value of `\frac{dy}{dx}` at `x=\frac{1}{2}` is `-\frac{1}{\sqrt{3}}`.

Question No: #130

if `y=tan^{-1}(\frac{\sqrt{1+sinx}-\sqrt{1-sinx}}{\sqrt{1+sinx}+\sqrt{1-sinx}})` where `0\lt\x\lt\frac{\pi}{2}`, then find the value of `\frac{dy}{dx}`