cosh(-x) = cosh x (This shows that cosh is an even function.) cosh(-x) = 1 / 2[ + e-(-x)] = 1 / 2( + ex) = Show transcribed image text There are 2 steps to solve this one.

2016年2月17日 · If we wanted to find, for example, the taylor series of #cosh(x)# around #x=0# then we set #x_0=0# and use the above definition. It is best to lay out two columns, one with the derivative and the other evaluating the value of #f^n(x_0)# at the point we wish to expand around. #f(x) = cosh(x)# #f(0) =1# #f'(x) = sinh(x)# #f'(0)=0# #f''(x) = cosh ...

2014年9月25日 · How do you find the Maclaurin series of #f(x)=sin(x)# ? How do you use a Maclaurin series to find the derivative of a function? How do I obtain the Maclaurin series for #f(x)= 2xln(1+x3)#?

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 …

2016年1月20日 · \\frac{d}{dx}(cosh(x))=sinh(x) Given cosh(x)=\\frac{e^x+e^(-x)}{2} Differentiating the right hand side of the equation with respect to x \\frac{d}{dx}(e^x)+\\frac{d ...

2015年11月7日 · What is #cosh(ln(x))#? Algebra Exponents and Exponential Functions Applications of Exponential Functions ...

2016年5月4日 · see below Let y=cosh^-1x then by definition x=cosh y =(e^y+e^-y)/2 2x=e^y+e^-y e^y-2x+e^-y=0 e^y-2x+1/e^y=0 e^(2y)-2xe^y+1=0 Let u=e^y then we have u^2-2xu+1=0--> Now use quadratic formula to solve u=(2x+-sqrt(4x^2-4))/2 e^y = (2x+-sqrt(4x^2-4))/2 e^y = (2x+-sqrt(4(x^2-1)))/2 2e^y = 2x+-2sqrt(x^2-1) e^y=x+-sqrt(x^2-1) ln e^y = ln (x+-sqrt(x^2-1 ...

Question: Prove the identity. cosh(x) + sinh(x) = ex First, use the definitions of the hyperbolic ...

2015年10月8日 · All Solutions for x: \(\begin{array}{rcl} \cosh{(x)} &=& \dfrac { e^x + e^{-x} } {2} = 0 \\ \dfrac { e^x + e^{-x} } {2} &=& 0 \\ e^x + e^{-x} &=& 0 \\ e^x + \dfrac{ 1 ...