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What is the second derivitive of sec 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.
What is tan x differentiated?
The derivative of ( \tan x ) with respect to ( x ) is ( \sec^2 x ). This can be derived using the quotient rule, since ( \tan x = \frac{\sin x}{\cos x} ), or recognized as a standard derivative. Thus, when differentiating ( \tan x ), you get ( \frac{d}{dx}(\tan x) = \sec^2 x ).
What is the anti derivative of sec?
The antiderivative of sec(x) is given by the formula: [ \int \sec(x) , dx = \ln | \sec(x) + \tan(x) | + C ] where ( C ) is the constant of integration. This result can be derived using a clever manipulation involving multiplying by a specific form of 1.
What is the second derivative of ln(tan(x))?
f'(x) = 1/tan(x) * sec^2(x) where * means multiply and ^ means to the power of. = cot(x) * sec^2(x) f''(x) = f'(cot(x)*sec^2(x) + cot(x)*f'[sec^2(x)] = -csc^2(x)*sec^2(x) + cot(x)*2tan(x)sec^2(x) = sec^2(x) [cot(x)-csc^2(x)] +2tan(x)cot(x) = sec^2(x) [cot(x)-csc^2(x)] +2
How do you prove that the derivative of sec x is equal to sec x tan x?
Show that sec'x = d/dx (sec x) = sec x tan x. First, take note that sec x = 1/cos x; d sin x = cos x dx; d cos x = -sin x dx; and d log u = du/u. From the last, we have du = u d log u. Then, letting u = sec x, we have, d sec x = sec x d log sec x; and d log sec x = d log ( 1 / cos x ) = -d log cos x = d ( -cos x ) / cos x = sin x dx / cos x = tan x dx. Thence, d sec x = sec x tan x dx, and sec' x = sec x tan x, which is what we set out to show.
What is the derivative of secant x?
The derivative of sec(x) is sec(x) tan(x).
What is the second derivitive of sec 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.
What is the derivative of 5secx?
You can take out any constant from a derivative. In other words, this is the same as 5 times the derivative of sec x.
What is the derivative of y equals sec x?
sec(x)tan(x)
What is the derivative of sec squared x?
derivative of sec2(x)=2tan(x)sec2(x)
What is the derivative of sec5X?
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.
What is the derivative of secxtanx?
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)
What is tan x differentiated?
The derivative of ( \tan x ) with respect to ( x ) is ( \sec^2 x ). This can be derived using the quotient rule, since ( \tan x = \frac{\sin x}{\cos x} ), or recognized as a standard derivative. Thus, when differentiating ( \tan x ), you get ( \frac{d}{dx}(\tan x) = \sec^2 x ).
What is the anti derivative of sec?
The antiderivative of sec(x) is given by the formula: [ \int \sec(x) , dx = \ln | \sec(x) + \tan(x) | + C ] where ( C ) is the constant of integration. This result can be derived using a clever manipulation involving multiplying by a specific form of 1.
What is the second derivative of ln(tan(x))?
f'(x) = 1/tan(x) * sec^2(x) where * means multiply and ^ means to the power of. = cot(x) * sec^2(x) f''(x) = f'(cot(x)*sec^2(x) + cot(x)*f'[sec^2(x)] = -csc^2(x)*sec^2(x) + cot(x)*2tan(x)sec^2(x) = sec^2(x) [cot(x)-csc^2(x)] +2tan(x)cot(x) = sec^2(x) [cot(x)-csc^2(x)] +2
How do you prove that the derivative of sec x is equal to sec x tan x?
Show that sec'x = d/dx (sec x) = sec x tan x. First, take note that sec x = 1/cos x; d sin x = cos x dx; d cos x = -sin x dx; and d log u = du/u. From the last, we have du = u d log u. Then, letting u = sec x, we have, d sec x = sec x d log sec x; and d log sec x = d log ( 1 / cos x ) = -d log cos x = d ( -cos x ) / cos x = sin x dx / cos x = tan x dx. Thence, d sec x = sec x tan x dx, and sec' x = sec x tan x, which is what we set out to show.
What is the derivative of tangent squared?
f(x)= tan2(x) f'(x)= 2tan(x)*sec2(x)
