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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 else can I help you with?
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 derivative of secant 2X?
d/dx sec(2x) = 2sec(2x)tan(2x)
What is the derivative of y equals tan x?
Given y = tan x: dy/dx = sec^2 x(secant of x squared)
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.
How do you identify sec x sin x equals tan x?
Rewrite sec x as 1/cos x. Then, sec x sin x = (1/cos x)(sin x) = sin x/cos x. By definition, this is equal to tan x.
What is the derivative of secant x?
The derivative of sec(x) is sec(x) tan(x).
What is the derivative of y equals sec x?
sec(x)tan(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 derivative of secant 2X?
d/dx sec(2x) = 2sec(2x)tan(2x)
What is the derivative of tangent?
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.
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)
How do I prove that sec A tan A sin A csc2 A?
It helps to convert everything to sines and cosines. Then you can often do lots of simplifications.Reminder: sec A = 1 / cos A tan A = sin A / cos A cosec A = 1 / sin A If you are unsure whether the two actually ARE equal, try evaluating both expressions (left and right) for some arbitrary value (for example, 10 degrees). If the two are NOT equal, then the expression are of course NOT equal. But if they ARE equal, you still need to prove that - since the claim is basically that the expressions are equal for ALL values of the variable.
What is the derivative of y equals tan x?
Given y = tan x: dy/dx = sec^2 x(secant of x squared)
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.
Can you help me prove math trig identities?
these are the identities i need sinΘcosΘ=cosΘ sec^4Θ-tan^4Θ=sec²Θcsc²Θ (1+sec²Θ)/(1-secΘ)=(cosΘ-1)/(cosΘ)
What is the derivative of tan x?
The derivative of tan(x) is sec2(x).(Which is the same as 1/cos2(x).
Prove that tan x equals tan 6x?
It is NOT equal. Try calculating tan x, and tan 6x, for a few values of "x", on your scientific calculator. Perhaps you are supposed to solve an equation, and see FOR WHAT values of "x" the two are equal?
