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for part a) let y=xcos(X) , the ln(y)=ln(xcos(X))=cos(x)ln(x), thus d/dx (ln(y(x)) = d/dx (cos(x)ln(x)), 1/y*dy/dx=cox(x)/x - sinxlnx => solve for dy/dx =>...
The point of intersection is where the lines would cross if we drew them on the same graph, in your exam you may be asked to find the coordinates of this point.
To do this we would first draw a ske...
Integrate ln(x) by parts with u=ln(x) and dv/dx=1 Gives x+xln(x)-x+c Therefore xln(x)+c
x2+5x+6=0 (x+3)(x+2)=0 x=-3 and x=-2
integral of 4x3 with respect to x =x4+C
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