2016年12月17日 · Calculus Differentiating Trigonometric Functions Derivatives of y=sec(x), y=cot(x), y= csc(x) 1 Answer

The derivatives of \sec(x), \cot(x), and \csc(x) can be calculated by using the quotient rule of differentiation together with the identities \sec(x)=\frac{1}{\cos(x ...

2018年4月13日 · #csc^2(x)=1/sin^2(x)# #d/dx[csc^2(x)]=d/dx[1/sin^2(x)]# #d/dx[1/sin^2(x)]=d/dx[[sin(x)]^{-2}]# let #u=sinx#. #d/dx[[sin(x)]^{-2}]=d/{du}[u^{-2}]d/dx[sinx]#

2015年3月15日 · The antiderivative of csc^2x is -cotx+C. Why? Before trying anything 'fancy' (subsitutuion, parts, trig sub, misc sub, partial fractions, et c.) try 'staightaway' antidifferentiation. Do you know a finction whose derivative is csc^2x? Go through the list: d/(dx)(sinx)=cosx d/(dx)(cosx)=-sinx d/(dx)(tanx)=sec^2x Hang on! that's good! The derivative if a co function …

2014年8月30日 · By Product Rule, we can find: #y'=-csc(x)cot^2(x)-csc^3(x)# Remember: #[scs(x)]'=-csc(x)cot(x)# #[cot(x)]'=-csc^2(x)#

2014年8月6日 · dy/dx = -1/sqrt(x^4 - x^2) Process: 1.) y = "arccsc"(x) First we will rewrite the equation in a form that is easier to work with. Take the cosecant of both sides: 2.) csc y = x Rewrite in terms of sine: 3.) 1/siny = x Solve for y: 4.) 1 = xsin y 5.) 1/x = sin y 6.) y = arcsin (1/x) Now, taking the derivative should be easier. It's now just a matter of chain rule. We know that …

2015年3月5日 · How do I find the antiderivative of #y=csc(x)cot(x)#? Calculus Introduction to Integration Integrals of Trigonometric Functions. 1 Answer

2016年6月5日 · To find the derivative of secant, we could either use the limit definition of the derivative (which would take a very long time) or the definition of secant itself:

2016年8月22日 · We will use the chain rule to differentiate. Let y = lnu and u = cscx The derivative of cscx, by the quotient rule, is (cscx)' = (-cosx)/sin^2x = -csc^2xcosx = -cotxcscx The derivative of lnu is 1/u. dy/dx= 1/u xx -cotxcscx = 1/cscx xx -cotxcscx = sinx xx (-cotx xx 1/sinx)=- cotx Hopefully this helps!

2016年11月10日 · #int cscx dx = int cscx/1*(cscx+cotx)/(cscx+cotx) dx# # = int (csc^2x+cscxcotx)/(cscx+cotx) dx# Let #u = cscx+cotx#, the #du = -(csc^2x+cscxcotx)#