If the x value changes u, then the y value changes by 7u.
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x5lnx?d/dx (uv)=u*dv/dx+v*du/dxd/dx (x5lnx)=x5*[d/dx(lnx)]+lnx*[d/dx(x5)]-The derivative of lnx is:d/dx(lnu)=(1/u)*[d/dx(u)]d/dx(lnx)=(1/x)*[d/dx(x)]d/dx(lnx)=(1/x)*[1]d/dx(lnx)=(1/x)-The derivative of x5 is:d/dx (xn)=nxn-1d/dx (x5)=5x5-1d/dx (x5)=5x4d/dx (x5lnx)=x5*[1/x]+lnx*[5x4]d/dx (x5lnx)=[x5/x]+5x4lnxd/dx (x5lnx)=x4+5x4lnx
By x3 I assume that you mean x3. In which case f(x)=x3-2x+1, and f'(x)=3x2-2. Therefore our iteration formula is: xn+1=xn- (xn3-2xn+1)/(3xn2-2) Starting with x0=0 we get: x1=0.5 x2=0.6 x3=0.617391304 x4=0.618033095 x5=0.618033988 x6=0.618033988 Starting with x0=0.9 we get: x1=1.065116279 x2=1.009457333 x3=1.000255451 x4=1.000000195 x5=1 x6=1 Starting with x0=-1.5 we get: x1=-1.631578947 x2=-1.618183589 x3=-1.618034007 x4=-1.618033989 x5=-1.618033989 The 3 real roots to f(x) are x=-1.618033989, x=0.618033988, and x=1
The integral of x4dx = x5/5 Evaluate that integral from 2 to 4 = 45/5 - 25/5 =(45 - 25)/5 = (1024-32)/5 = 992/5 = 198.4
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The answer to this depends on what you mean by "x 7" If you mean: x2 -7x, then it can be factored out as x(x - 7) If you mean: x2 - x7, then you can factor it out as: x2(1 - x5) If you mean: x2 - x + 7, then it can not be factored If you mean: (x2 - x)7, then the inner term can be factored, giving you (x[x - 1])7 If you mean something else, then you will need to be more clear with your question.