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Firstly I would state the rule that for any composite function fg, its domain is always the domain of g. If the student would like to know why this is the case, I would simplify the expression fg(x) to f(...

We start off by differentiating the equation implicitly which will give us:2(x-y) -2(x-y)dy/dx = 6 + 5dy/dxThen we rearrange to get dy/dx on it's own:dy/dx = (2x-2y-6)/(2x-2y+5)
For the second part o...

Using integration by parts by letting u=x and dv/dx=e^-x. this implies that du/dx=1 and v=-e^-xThe By Parts formulae is u.v - integral(v.du/dx) = -xe^-x - in...

Since we differentiate a function to find the gradient of a curve at any point, we need to reverse that to find the equation of the curve. We do this by integrating with respect to x:If you have a constan...

Formula for a straight line y-y₁=m(x-x₁), where m is the gradient substituting in the values given to find the gradient we get 4-1=m(3+2), therefore m= 3/5the midpoint of the two poi...