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1/xlnx Use the chain rule: ln(ln(x)) The derivative of the outside is1/ln(x) times the derivative of the inside. 1/[x*ln(x)]
2 ln(9) + 2 ln(5) = 2 ln(x) - 3ln(81) + ln(25) = ln(x2) - 37.61332 = ln(x2) - 3ln(x2) = 10.61332ln(x) = 5.30666x = e5.30666 = 201.676 (rounded)
-1/(2*x2)
If you mean: y =(lnx)3 then: dy/dx = [3(lnx)2]/x ddy/dx = [(6lnx / x) - 3(lnx)2] / x2 If you mean: y = ln(x3) Then: dy/dx = 3x2/x3 = 3/x = 3x-1 ddy/dx = -3x-2 = -3/x2
There are several steps involved in how one can solve the derivative x plus y - 1 equals x2 plus y2. The final answer to this math problem is y'(x) = (1-2 x)/(2 y-1).