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Take a right angled triangle with hypotenuse of length 1, and angle at the bottom of the hypotenuse equal to x. We will let o denote the length of the side opposite the angle, and a denote the length of t...

The first step is to find dx/dy in terms of y, which when differentiating comes out as 3/2cos(y/2), so dy/dx in terms of y is the reciprocal of this.The next step is to eliminate the y dependent terms, wh...

dy/dx = 6x^1/2 + 2 y = (62/3)x^3/2 + 2x + c = 4x^3/2 + 2x + c 38 = (416^3/2) + (2*16) + c c = 38 - 256 - 32 = -250 y = 4x^3/2 + 2x - 250

A tricky question which is actually really simple if you know the technique and will save a lot of time during the exam. The method is to simply add 1 and minus one to the numerator. so the integral becom...

Using the formula for differentiation: if y=ax^N then dy/dx= Na*x^N-1 apply this to each term in the equation given; first term: dy/dx= 9x²second term: dy/dx= 14xth...