y = e^ln x using the fact that e to the ln x is just x, and the derivative of x is 1:
y = x
y' = 1
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If the function is (ln x)2, then the chain rules gives us the derivative 2ln(x)/x, with the x in the denominator. If the function is ln (x2), then the chain rule gives us the derivative 2/x.
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)
e^(-2x) * -2 The derivative of e^F(x) is e^F(x) times the derivative of F(x)
The derivative of ln(10) is 1/10. This is because the derivative of the natural logarithm function ln(x) is 1/x. Therefore, when differentiating ln(10), the derivative is 1/10.
The derivative of ln x is 1/x The derivative of 2ln x is 2(1/x) = 2/x