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 sum rule.

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) = e^0=1 the Mc Laurin series for e^x is: e^x=sum_(n=0)^oo x^n/(n!) Now as: sinhx = (e^x-e^(-x))/2 We have: sinhx = 1/2[sum_(n=0)^oo x^n/(n!)-sum_(n=0)^oo (-x)^n/(n!)] and it is easy to see that for n even the terms are the same ...

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 #sinh(x)#, so we should work out the #M#-bound for #cosh(x)# The hyperbolic cosine function is always increasing, so the largest value on the interval will be at #1 \/ 2#:

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 we rewrite #sinhx=(e^x-e^-x)/2# as...

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 functions are:

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: =cosh x + sinh x =(e^x+e^-x)/2 +(e^x-e^-x)/2 =(e^x+e^-x +e^x-e^-x)/2 =(e^x +e ...

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 ...

2018年5月12日 · 12996 views around the world You can reuse this answer ...

2014年12月19日 · The definition of cosh(x) is (e^x + e^-x)/2, so let's take the derivative of that: d/dx ((e^x + e^-x)/2) We can bring 1/2 upfront. 1/2(d/dxe^x + d/dxe^-x) For the first part, we can just use the fact that the derivative of e^x = e^x: 1/2(e^x + d/dxe^-x) For the second part, we can use the same definition, but we also have to use the chain rule. For this, we need the derivative of …

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. Answer link