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Using the product rule we know that dy/dx = uv' + vu' where u = 3x^3; v = e^x. e^x differentiates to itself multiplied by any number in front of the x. u' = 9x^2; v' = e^x. Therefore dy/dx = 3x^3e...
when integrating with the limits, you substitute in the two limits into the integral and then take them away. As both will have added the constant 'c', you are taking c away from c and so cancel out, so i...
F(x+2) is simply F(x) but whenever you see an 'x' replace it with 'x+2'. So when x is say 3, F(x) is F(3) and F(x+2) is F(5). If we draw this out on a graph, we see that this has the effect of shifting th...
The minimum point occurs where dy/dx=0
We have 2 options: 1.) Expanding the brackets 2.) The product rule of differentiation
The shortest is the product rule: dy/dx= (d/dx)(3-x).(x2ZEAnswered by Zita E. • Maths tutor2782 Views
Note: time, t, is measured in minutes, and height, h, is measured in metres.
Let k>0, a constant.
The differential equation to be solved is given by: dh/dt = - k(h)^0.5.
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