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1. You first find the gradient of the tangent to the curve at this given co-ordinate by differentiating the given equation of the curve, and then, assuming the equation of the curve is in terms of x, r...

Gradient of line 3x+5y=7: 5y=-3x+7, y=-3/5x+7/5 gradient = -3/5 Gradient of perpendicular line: 5/3 Perpendicular line with points: y+3=5/3(x+2), 3y+9=5(x+2), 3y+9=5x+10, 5x-3y+1=0

Using the chain rule, we let u = sin(x) and v = x^2. Then dy/dx = udv/dx + vdu/dx. dv/dx = 2x and du/dx = cos(x). So dy/dx = sin(x)2x + x^2cos(x).

=(5a x 3a) + (5a x -4b) + (3a x -2b) + (-2b x -4b) =15a^2 - 20ab - 6ab + 8b^2 =15a^2 - 26ab + 8b^2

Let y = arctan(x). Then x = tan(y).

Differentiate using the chain rule and rearrange: d(x)/dx = d(tany)/dx So 1 = sec^2(y) * dy/dx dy/dx = 1/sec^2(y)

But from identity sin^2(y) + cos^2(y) = ...