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x=1/3 and y=6 or x=-2 and y=-1

Have a picture of full working with annotation to go through during interview.Here is rough outline:y = cosh^-1(x)x = cosh(y)x = (e^y+e^-y)/22x = e^y+e^-y...

let's start by defining y = arsinh(cos(x)). taking sinh of both sides gives sinhy = cosx. (since sinh(arsinhz) = z). Now we can differentiate both sides wrt x. The RHS differentiates to -sinx. We can use ...

Express 6 in the exponential form of (6*exp(2πin)), where n is an integertake the fourth root of this, remembering that taking the fourth root is taking to the power of 1/4, and by rules of indices the ex...

Firstly, I haven't seen the notation I used in alevel but I just used it for the sake of ease of typing it online.1st. Sub in the trial solution f(x)= Ae^(mx) and its derivatives- f'(x)= Ame^(mx) and f''(...