d/dx csc(x) = - csc(x) tan(x)
∫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
The derivative of ln x is 1/x The derivative of 2ln x is 2(1/x) = 2/x
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).
derivative of sec2(x)=2tan(x)sec2(x)
d/dx csc(x) = - csc(x) tan(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.
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d/dx (-cscx-sinx)=cscxcotx-cosx
negative cotangent -- dcot(x)/dx=-csc^2(x)
∫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
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).
tan(x)*csc(x) = sec(x)
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!
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)
They are co-functions meaning that 90 - sec x = csc x.
It's easiest to show all of the work (explanations/identities), and x represents theta. cosxcotx + sinx = cscx cosx times cosx/sinx + sinx = csc x (Quotient Identity) cosx2 /sinx + sinx = csc x (multiplied) 1-sinx2/sinx + sinx = csc x (Pythagorean Identity) 1/sinx - sinx2/sinx + sinx = csc x (seperate fraction) 1/sinx -sinx + sinx = csc x (canceled) 1/sinx = csc x (cancelled) csc x =csc x (Reciprocal Identity)