How do you differentiate x^x?

To differentiate xx, we first let y = xx. (Note that xx is not in the form xc where c is a constant or ax where a is a constant so the usual differentiation formulas cannot be used). The trick here is to take the natutral logarithm of both sides. Then you obtain, ln(y) = ln(xx). From here you need to use the rule that ln(xx) = xln(x). So currently we have ln(y) = xln(x). From here we can differentiate implicitly to get: 1/y multiplied by dy/dx = ln(x) + 1 (differentiate right hand side using product rule and left hand side using chain rule).The final step is to multiply through by y and substitute xx back in for y. This gives you: dy/dx = xx(ln(x) + 1).  

AP

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(a) By using a suitable trigonometrical identity, solve the equation tan(2x-π/6)^2 =11-sec(2x-π/6)giving all values of x in radians to two decimal places in the interval 0<=x <=π .


Differentiate: (12x^3)+ 4x + 7


Find the set of values for which: x^2 - 3x - 18 > 0


How would I answer this question? Use factor theorem to show (x-2) is a factor of f(x) = 2x^3 -7x^2 +4x +4.