ln(x)+1. This can be found using the product rule: if f(x)=uv, f'(x)=u'v+v'u, so if u is x and v is ln(x), the product rule gives (1)(ln(x))+(x)(1/x)=ln(x)+1.
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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
I get x*x^x-1 + lnx*x^x = x^x + x^xlnx = x^x * (1+lnx) Here, ^ is power; * = times; ln = natural logratithm ( base e)
A dot A = A2 do a derivative of both sides derivative (A) dot A + A dot derivative(A) =0 2(derivative (A) dot A)=0 (derivative (A) dot A)=0 A * derivative (A) * cos (theta) =0 => theta =90 A and derivative (A) are perpendicular
The derivative of e7x is e7 or 7e.The derivative of e7x is 7e7xThe derivative of e7x is e7xln(7)
the derivative is 0. the derivative of a constant is always 0.