Derivative of Tan x

Derivative of Tan x Derivative of Tan x The derivative of tan(X) Since tan(X)=sin(X)/cos(X), we have sin(X) as the function u(X) and cos(X) as the function v(X). Putting these into the formula d[uv]/dX=(vdu/dX - udv/dX)/v2 we get d[tan(X)]/dX = (cos(X)cos(X) + sin(X)sin(X))/cos2(X) But on the top we have sin2(X)+ cos2(X), which is always 1. So our result simplifies to d[tan(X)]/dX = 1/cos2(X).

But that is sec2(X), since sec(X)=1/cos(X).

So the derivative of tan(X) is sec2(X).

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This is the result we had in the earlier section on differentiating trig funcions.

Derivative of 1/(sec x - tan x)? lets simplify your equation..........

1/(sec x - tan x) =(sec x + tan x) / (sec x - tan x)(sec x + tan x) =(sec x + tan x) / (sec^2 - tan^2) =(sec x + tan x) ............. (since sec^2 = 1 + tan^2 ) now derivatin above simplified equation........ sec x* tan x + sec^2 = sec x ( sec x + tan x )....... ans.

Proving the derivative of tan x?

Tan' (x)= (sec x)^2 lim (h→0) [f(x + h) - f(x)]/h = lim (h→0) [tan(x + h) - tan(x)]/h = lim (h→0) [sin(x + h)/cos(x + h) - sin(x)/cos(x)]/h = lim (h→0) [(cos(x)sin(x + h) - sin(x)cos(x + h))] / [h cos(x)cos(x + h)] = lim (h→0) [cos(x)(sin(x)cos(h) + cos(x)sin(h)) - sin(x)(cos(x)cos(h) - sin(x)sin(h))] / [h cos(x)cos(x + h)] = lim (h→0) [sin(x)cos(x)cos(h) + cos²(x)sin(h) - sin(x)cos(x)cos(h) + sin²(x)sin(h)) ] / Read More About Properties of square Roots

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[h cos(x)cos(x + h)] = lim (h→0) [cos²(x)sin(h) + sin²(x)sin(h)) ] / [h cos(x)cos(x + h)] = lim (h→0) [sin(h)(cos²(x) + sin²(x))] / [h cos(x)cos(x + h)] = lim (h→0) sin(h) / [h cos(x)cos(x + h)] = lim (h→0) [sin(h)/h] * lim (h→0) [1/cos(x)cos(x + h)] = 1 * 1/ [cos(x)cos(x + 0)] = 1/cos²(x) = sec²(x)

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