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What is the derivative of x5lnx?
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
Real root of x3-2x plus 1 in newton raphson method?
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
Evaluate the integral x4 dx from 4 to 2?
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
What is the solution set of the inequality 7 x5?
Unfortunately, limitations of the browser used by Answers.com means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. Please resubmit your question spelling out the symbols as "plus", "minus", "times", "equals", and inequlity symbols. All that is visible here are two expressions separated by a space: no inequality sign at all!
How do you factorise x squared minus x 7?
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.
