2014年12月12日 · (d(sinh(x)))/dx = cosh(x) Proof: It is helpful to note that sinh(x):=(e^x-e^-x)/2 and cosh(x):=(e^x+e^-x)/2. We can differentiate from here using either the quotient rule or the …

2017年12月31日 · #sqrt(x-3)=y# #larr# This is the inverse function denoted by as #f^(-1)(x)# Given #sinhx=(e^x-e^-x)/2# , finding it's inverse is the same concept: Suppose #y=sinhx# , so …

2016年12月26日 · sinhx =sum_(k=0)^oo x^(2k+1)/((2k+1)!) We can derive the McLaurin series for sinh(x) from the one othe exponential function: as for every n: [(d^n)/(dx^n) e^x ]_(x=0) = …

2018年2月10日 · The higher order derivatives of #sinh(x)# will either be #sinh(x)# or #cosh(x)#. If we consider the definitions for them, we see that #cosh(x)# will always be larger than …

2018年4月21日 · R=oo Let's first find the Maclaurin series expansion for sinhx: f(x)=sinhx=(e^x-e^-x)/2, f(0)=(e^0-e^0)/2=0 f'(x)=coshx=(e^x+e^-x)/2, f'(0)=(e^0+e^0)/2=1 f''(x)=sinhx ...

2016年5月4日 · see below Use the definition cosh x = (e^x+e^-x)/2 and sinh x = (e^x-e^-x)/2 Left Side: [(e^x+e^-x)/2 + (e^x-e^-x)/2]^n =[(e^x+e^-x + e^x-e^-x)/2]^n =[(2e^x)/2]^n =e ...

2016年9月18日 · sinhx=+-12/5, coshx=+-13/5, cothx=13/12, sechx=+-5/13 and cschx=+-5/12 We can use relations between hyperbolic functions to find them.

2016年10月24日 · It's not really an attempt to treat "sinh" as a word, more a convention to convey that it's a hyperbolic version of "sine" Explanation: The full names of the six trigonometric …

2018年2月11日 · #ln(1+cosh(x))+C#, which is our final answer. We remove the absolute value from the logarithm because we note that #cosh# is positive on its domain so it's not necessary. …

f(x) = 2x + (2x^3)/3 + x^5/(20) + x^7/(630) + ... \ \ \ \ x in RR There are several ways to develop a solution, we could start from first principles and form the Taylor series for the hyperbolic …