Dec 19, 2014 · 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 …

$\cosh x = \cos ix$ $\sinh x = i \sin ix$ which, IMO, conveys intuition that any fact about the circular functions can be translated into an analogous fact about hyperbolic functions. e.g. I know that there is a double-angle formula for $\cos$. Therefore, there should be a similar double-angle formula for $\cosh$.

Jan 20, 2016 · \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}{dx ...

What's wrong with my differentiation (help finding a derivative)? 1 Derivative of a function with hyperbolic cosine and exponent $\frac{e^{4x}}{x^3 \cosh (2x)}$

Apr 8, 2018 · But when I attempted the question, I tried to convert $\cosh^2(2x)$ into $\frac{\cosh(4x)+1}2$, using the identity $\cosh(2x)=2\cosh^2(x)-1$. After the conversion, the answer I get differentiating this will be $2\sinh(4x)$ which is a different answer?

Dec 30, 2016 · The answer is =1/(2sqrtxsqrt(x-1)) We need (sqrtx)'=1/(2sqrtx) (coshx)'=sinhx cosh^2x-sinh^2x=1 Here, we have y=cosh^(-1)(sqrtx) Therefore, coshy=sqrtx Taking the ...

Answer to: Find the derivative of f(x) = e^x cosh x. Simplify where possible. By signing up, you'll get thousands of step-by-step solutions to your...

The functions $\cosh$ and $\sinh$ are known as hyperbolic functions.The definitions are: $$\cosh x = \frac{e^x + e^{-x}}{2} \qquad \quad \sinh x = \frac{e^x - e^{-x}}{2} $$ It is easy to remember the signs, thinking that $\cos$ is an even function, and $\sin$ is odd.

Oct 1, 2017 · y=cosh^(-1) (sinhx) coshy=sinhx y'*sinhy=coshx y'=coshx/sinhy y'=coshx/sqrt[(coshy)^2-1] y'=coshx/sqrt[(sinhx)^2-1] 1) I transformed y=cosh^(-1) (sinhx) into coshy=sinhx. 2) I took differentiation both sides. 3) I left y' alone dividing both sides by sinhy. 4) I replaced sinhy with hyperbolic function in terms of x using sinhy=sqrt[(coshy)^2 …

The derivative of $\cosh(x)$ is $\sinh(x)$ and the derivative of $\sinh(x)$ is $\cosh(x)$, so you're ...