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Say arctan of x is equal to a value y. Now take the tangent of both sides; x now equals tan of y! Easy from here, differentiate both sides wrt x. Now 1 equals sec^2y dy/dx, and you can rearrange to find d...

The line of intersection will be parallel to both planes. We can write the equations of the two planes in 'normal form' as r.(2,1,-1)=4 and r.(3,5,2)=13 respectively. We can then read off the normal vecto...

Consider the case n=1. Then 1(1+1)(2*1+1)/6 = 1 = 1^2 and so the claim is true for n=1. Suppose the claim is true for some positive integer n, so that 1+4+9+...+n^2 = n(n+1)(2n+1)/6. Then by the inductive...

The general method is: 1)write down what needs to be shown (the claim) 2)check it holds for the lowest value of n required (normally n=1 but check question) 3)write down sentence: 'Suppose when n=m the cl...

The first step in tackling this problem is to find the intersection points of the graph with the x and y axes. The y-intercept can be found by substituting x=0 into the function, so y=(0-1)/(0+1)=-1 and t...