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Simplify e to the 1 plus Inx power?
e1 + (lnx) = e1 * e(lnx) = e * x = ex
What is the derivative of 1 lnx?
The derivative of 1/lnx, can be found easily using either the chain rule or the quotient rule. It is -1/[x*(lnx)2]
What is the integral of x.lnx?
Do you want ∫x lnx dx? Let's call that I, which we now seek to find. The solution is I = ½ x2 lnx - ¼ x2; here is how we can find it: Let z = lnx. Then, x = ez, dx = ez dz, and dI = x lnx dx. Then, dI = (ez)(z)(ez dz) = ze2z dz = ¼ (2z e2z d(2z)). Thus, 4dI = wew dw, where we let w = 2z = 2 lnx. Now, d(wew) = ew dw + w d(ew) = ew dw + wew dw = ew dw + 4dI; hence, 4dI = d(wew) - ewdw = d(wew) - d(ew)­ = d((w - 1)ew). Therefore, 4I = (w - 1)ew = (2 lnx - 1)x2 = 2x2 lnx - x2; and I = ½ x2 lnx - ¼ x2, which is the answer we sought. Checking, we differentiate back, to confirm that I' = x lnx: I = ¼ x2(2 lnx - 1), whence, 4I' = (x2(2 lnx - 1))' = 2x(2 lnx - 1) + x2 (2/x) = 2x(2 lnx - 1) + 2x = 2x(2 lnx) = 4x lnx; thus, I' = x lnx, re-assuring us that we have integrated correctly.
What is the derivative of lnx squared?
I do not see why the chain rule would not work here. d/dx (inx)^2 = 2(lnx) * 1/x = 2(lnx)/x
What is the Derivative of x to the power of x?
The solution to this is: (xx)'= (elnx to the power of x)'= (exlnx)'= (xlnx)'*exlnx= [x(1/x) + 1(lnx)]*exlnx = (lnx+1)*exlnx= (lnx+1)*xx
