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We can identify xe^(-2x) as a product and hence we will most probably need to use integration by parts.
We then set u = x and v' = x^(-2x). It is important to do it this way round so that when we d...
The rules to know are 1) when differentiating e to the power of x... be it 2x or 100x... you bring down the number in front of x, and leave the power as it is. in our case e^(2x) goes to 2e^(2x).
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).
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