You can take out any constant from a derivative. In other words, this is the same as 5 times the derivative of sec x.
the derivative of 3x is 3 the derivative of x cubed is 3 times x squared
2xsin2x+2x2sinxcosx
The idea is to use the chain rule. Look up the derivative of sec x, and just replace "x" with "5x". Then multiply that with the derivative of 5x.
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 derivative of sec(x) is sec(x) tan(x).
Write sec x as a function of sines and cosines (in this case, sec x = 1 / cos x). Then use the division formula to take the first derivative. Take the derivative of the first derivative to get the second derivative. Reminder: the derivative of sin x is cos x; the derivative of cos x is - sin x.
Given y = tan x: dy/dx = sec^2 x(secant of x squared)
You can take out any constant from a derivative. In other words, this is the same as 5 times the derivative of sec x.
sec(x)tan(x)
f(x)= tan2(x) f'(x)= 2tan(x)*sec2(x)
the derivative of 3x is 3 the derivative of x cubed is 3 times x squared
10 x
x squared
Let k = 0 9x18 squared x 17 x 18 k is a constant. Its anti-derivative is kx + C, where C is a constant. The anti-derivative squared is (kx+ C) squared.
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, with respect to x, is -x/sqrt(1-x2)