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```
y= tan^-1(secx-tanx), then dy/dx

```
11 years ago 8 Points
```							d (tan-1(x))/dx = 1/(1+x^2)To prove the above result, let y=tan-1(x)or,tan(y)=x, differentiate w.r.t. x,sec^2(y).dy/dx=1, or, dy/dx=cos^2(y) ...(i)Let p=tan^-1(x), or x=tan(p), so cos(p)=1/(1+x^2)^0.5,So, cos(p)=cos(tan-1(x))=1/(1+x^2)^0.5.Therefore, dy/dx=cos^2(y)=cos^2(tan-1(x))=1/(1+x^2)^0.5In the original questiondy/dx=(sec(x)tan(x)-sec^2(x)) / (1 + (sec(x) - tan(x))^2)       =-1/2 (after simplification, using 1+tan^2(x)=sec^2(x))
```
11 years ago
```							derivative of arc tan x is 1/(1+x2)
here dy/dx= ( secx.tanx - sec2x) / (1+sec2x+tan2x-2.secx.tanx)
= secx(secx-tanx)/2.tanx(secx-tanx)
=secx/2tanx
=(cosec x) /2

--
regards
Ramesh
```
11 years ago
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