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`        differentiate tan^-1(1+ax/1-ax) with respect to sqrt(1+a^2x^2)`
5 years ago

Arun Kumar
IIT Delhi
256 Points
```							Hi$\\a=tan^{-1}(\frac{1+ax}{1-ax}) \\b=\sqrt{1+a^2x^2} \\\\This is NCERT level question \\\frac{da}{db}=\frac{\frac{da}{dx}}{\frac{db}{dx}}$Please evaluate yourself.Thanks & Regards, Arun Kumar, Btech,IIT Delhi, Askiitians Faculty
```
5 years ago
kantha mani
27 Points
```							Let x = (1/a) tan yThen, u = tan^1 [(1 + tan y)/(1 - tan y)]              = tan^1 [(tan pi/4 + tan y)/(1 - tan pi/4 tan y)]              = tan^1 [tan (pi/4 +y)]              = (pi/4 + y)              = pi/4 + tan^1(ax)du/dx = 1/(1+a^2x^2)v = (1 + a^2x^2)^1/2dv/dx = a^2x/ ( (1 + a^2x^2)^1/2du/du = 1/a^2x (1 + a^2x^2)^1/2
```
one year ago
kantha mani
27 Points
```							Let x = (1/a) tan yThen, u = tan^-1 [(1 + tan y)/(1 - tan y)]              = tan^-1 [(tan pi/4 + tan y)/(1 - tan pi/4 tan y)]              = tan^-1 [tan (pi/4 +y)]              = (pi/4 + y)              = pi/4 + tan^-1(ax)du/dx = 1/(1+a^2x^2)v = (1 + a^2x^2)^1/2dv/dx = a^2x/ ( (1 + a^2x^2)^1/2du/du = 1/a^2x (1 + a^2x^2)^1/2
```
one year ago
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• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions