sec(x)tan(x)
d/dx[ tan-1(x) ] = 1/(1 + x2)
The derivative with respect to 'x' of sin(pi x) ispi cos(pi x)
I am assuming the you are talking about the graph of the derivative. The graph of the derivative of F(x) is the graph such that, for any x, the value of x on the graph of the derivative of F(x) is the slope at point x in F(x).
the derivative of tangent dy/dx [ tan(u) ]= [sec^(2)u]u' this means that the derivative of tangent of u is secant squared u times the derivative of u.
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(arctan x) = X = 1/(1 + x2)
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)
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)
f(x)= tan2(x) f'(x)= 2tan(x)*sec2(x)
The derivative of ln x is 1/x The derivative of 2ln x is 2(1/x) = 2/x
The derivative of cos(x) is negative sin(x). Also, the derivative of sin(x) is cos(x).