The derivative of sec(x) is sec(x) tan(x).
The angle is the arc-tan of the gradient of the line. That is to say, the tangent of that angle is the gradient of the line or the angle between the straight line and the positive x-axis. Arc tan may also be written as tan-1 but that is frequently confused with 1/tan or the cotangent function.
Derivative of 1/x 1/x = x-1 Take the derivative (-1)x(-1-1) = -x-2 = 1/x2
cot2x-tan2x=(cot x -tan x)(cot x + tan x) =0 so either cot x - tan x = 0 or cot x + tan x =0 1) cot x = tan x => 1 / tan x = tan x => tan2x = 1 => tan x = 1 ou tan x = -1 x = pi/4 or x = -pi /4 2) cot x + tan x =0 => 1 / tan x = -tan x => tan2x = -1 if you know about complex number then infinity is the solution to this equation, if not there's no solution in real numbers.
The derivative of sin(x) is cos(x).
The derivative of tan(x) is sec2(x).(Which is the same as 1/cos2(x).
The derivative of sec(x) is sec(x) tan(x).
sec(x)tan(x)
d/dx csc(x) = - csc(x) tan(x)
Regardless of what 'x' is, (x)0 = 1 . tan(1 radian) = 1.55741 (rounded) tan(1 degree) = 0.01745 (rounded) We can't remember the derivative of the tangent right now, but it doesn't matter. This particular tangent is a constant, so its derivative is zero.
d/dx[ tan-1(x) ] = 1/(1 + x2)
The derivative of the natural log is 1/x, therefore the derivative is 1/cos(x). However, since the value of cos(x) is submitted within the natural log we must use the chain rule. Then, we multiply 1/cos(x) by the derivative of cos(x). We get the answer: -sin(x)/cos(x) which can be simplified into -tan(x).
Given y = tan x: dy/dx = sec^2 x(secant of x squared)
There is no difference in meaning between the two. It is usually spelled in lowercase, though (arc tan, or arctan).
d/dx(uv)=u*dv/dx+v*du/dxd/dx(secxtanx)=secx*[d/dx(tanx)]+tanx*[d/dx(secx)]-The derivative of tanx is:d/dx(tan u)=[sec(u)]2*d/dx(u)d/dx(tan x)=[sec(x)]2*d/dx(x)d/dx(tan x)=[sec(x)]2*(1)d/dx(tan x)=(sec(x))2=sec2(x)-The derivative of secx is:d/dx(sec u)=[sec(u)tan(u)]*d/dx(u)d/dx(sec x)=[sec(x)tan(x)]*d/dx(x)d/dx(sec x)=[sec(x)tan(x)]*(1)d/dx(sec x)=sec(x)tan(x)d/dx(secxtanx)=secx*[sec2(x)]+tanx*[sec(x)tan(x)]d/dx(secxtanx)=sec3(x)+sec(x)tan2(x)
The angle is the arc-tan of the gradient of the line. That is to say, the tangent of that angle is the gradient of the line or the angle between the straight line and the positive x-axis. Arc tan may also be written as tan-1 but that is frequently confused with 1/tan or the cotangent function.
f(x)= tan2(x) f'(x)= 2tan(x)*sec2(x)