0
What else can I help you with?
What is the derivative of cscx?
d/dx csc(x) = - csc(x) tan(x)
What is the anti derivative of cscxcotx?
∫cscxcotx*dx∫csc(u)cot(u)*du= -csc(u)+C, where C is the constant of integrationbecause d/dx(csc(u))=-[csc(u)cot(u)],so d/dx(-csc(u))=csc(u)cot(u).∫cscxcotx*dxLet:u=xdu/dx=1du=dx∫cscucotu*du= -csc(u)+CPlug in x for u.∫cscxcotx*dx= -csc(x)+C
What is the derivative of 2lnx?
The derivative of ln x is 1/x The derivative of 2ln x is 2(1/x) = 2/x
What is the derivative of 3cosx?
The derivative of 3cos(x) is -3sin(x). This can be found using the chain rule, which states that the derivative of a composition of functions is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. In this case, the derivative of cos(x) is -sin(x), and when multiplied by the constant 3, we get -3sin(x) as the derivative of 3cos(x).
What is the derivative of sec squared x?
derivative of sec2(x)=2tan(x)sec2(x)
What is the derivative of cscx?
d/dx csc(x) = - csc(x) tan(x)
What is the derivative of x-4cscx 2cotx?
To find the derivative of the function ( f(x) = x - 4 \csc(x) \cdot 2 \cot(x) ), we first differentiate each term separately. The derivative of ( x ) is ( 1 ). For the second term, we apply the product rule: the derivative of ( -4 \csc(x) \cdot 2 \cot(x) ) involves differentiating ( -4 \csc(x) ) and ( 2 \cot(x) ), resulting in ( -4(2(-\csc(x)\cot^2(x) - \csc^2(x))) ). Thus, the complete derivative is ( f'(x) = 1 - 4 \left( 2(-\csc(x)\cot^2(x) - \csc^2(x)) \right) ).
What is the anti-derivative of co secant x?
According to Wolfram Alpha, input:integral csc x it is -log[cot(x) + csc(x)] + constant You can verify this by taking the derivative of the purported integral.
What is the derivative of 3tanx-4cscx?
7
How do you find the derivative of - csc x - sin x?
d/dx (-cscx-sinx)=cscxcotx-cosx
What is the anti-derivative of secant squared?
negative cotangent -- dcot(x)/dx=-csc^2(x)
What is the derivitive of cenat?
The derivative of the function ( \csc(x) ) (cosecant) is given by ( -\csc(x) \cot(x) ). If you meant a different function by "cenat," please clarify, as "cenat" doesn't correspond to a standard mathematical term.
What is the anti derivative of cscxcotx?
∫cscxcotx*dx∫csc(u)cot(u)*du= -csc(u)+C, where C is the constant of integrationbecause d/dx(csc(u))=-[csc(u)cot(u)],so d/dx(-csc(u))=csc(u)cot(u).∫cscxcotx*dxLet:u=xdu/dx=1du=dx∫cscucotu*du= -csc(u)+CPlug in x for u.∫cscxcotx*dx= -csc(x)+C
Proof of the derivative of the cosecant function?
Express the cosecant in terms of sines and cosines; in this case, csc x = 1 / sin x. This can also be written as (sin x)-1. Remember that the derivative of sin x is cos x, and use either the formula for the derivative of a quotient (using the first expression), or the formula for the derivative of a power (using the second expression).
Can you use the equation given below to find the second derivative of pi divided by 6 if fx equals cscx?
pi divided by 6 is a constant and so its first derivative is 0. And since that is also a constant, the second derivative is 0. It is not clear what f(x) = csc(x) has to do with that!
What is tan x csc x?
tan(x)*csc(x) = sec(x)
How do you get the second derivative of g of x equals xcscx where x equals theta?
T=theta so that it will not look so messy. g(T)=TcscT To find the first derivative, you must use the product rule. Product rule is derivative of the first times the second, plus the first times the derivative of the second, which will give you: g'(T)=0xcscT + Tx-cscTcotT, which simplifies: g'(T)= -cscTxcotT Now, take the derivative of that to get the second derivatice. In order to do that, you have to do the product rule again. g"(T)=(cscTcotT)cotT + -cscT(-csc^2T) {that's csc squared} which simplifies: g"(T)= cscTcot^2(T) + csc^3 (T)
