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Firstly, Take equation 1 (2y+4x=14) and divide it all by 2. This gives y+2x=7. Next make Y the 'subject' AKA put it on it's own. To do this Subtract 2x from both sides. Now we have Y=7-2X. Here you can se...

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).