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1. 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)...

Differentiate to find dy/dx = 3x + 2;at point (-1,0) dy/dx = -1substitute in to y = mx + c, noting m = -1 and the line passes through (-1,0) yields c = -1y = -x - 1, simple to sketch this line.curve sketc...

When factorising isn't working, we can try using the quadratic formula. First, we need our quadratic equation and find a,b and c - the coefficients of x and the constant. Now, we can wri...

x = 30ºcos(30) = √3/2sec(30) = 2/√3 or 2√3/3

RHS: cosh(x)cosh(y) + sinh(x)sinh(y) = 1/4(e^x + e^-x)(e^y + e^-y) + 1/4(e^x - e^-x)(e^y - e^-y) = 1/4(e^x.e^y + e^x.e^-y + e^-x.e^y + e^-x.e^-y + e^x.e^y - e^x.e^-y - e^-x.e^y + e^-x.e^-y) = 1/4(2e^x.e^y...