FP2 (old specification) - How do you find the derivative of arsinhx?
- y = arsinhx We can rearrange this to be 2) sinhy = x, which makes this challenge slightly easier as we know that the derivative of sinhx is coshx and vice versa.Knowing the above we get coshy(dy/dx) = 1, we can rearrange this to get 3) dy/dx = 1/coshy.Next we have to think about our hyperbolic functions rules and find one that has both sinhx and coshx in it so that we can make an easy substitution. The hyperbolic function rule that we will use is 4) cosh2x - sinh2x = 1. If we square both sides of sinhy = x (eq 2), we will get sinh2y = x2. We can substitute this into the above equation (4) to get cosh 2x -x2=1, rearranging this we get cosh2x= 1+x2.Next we can see that we need to substitute cosh2x into our dy/dx equation. We can do this by squaring both sides of dy/dx = 1/coshx to get 5) (dy/dx)2= 1/cosh2x.Then substituting cosh2x = 1+x2 into the above equation (5) we get (dy/dx)2 = 1/(1+x2), we then should simplify this to get it in the form of dy/dx = 1/(1+x2)1/2.This is how you find the derivate of arsinhx.
