How do I differentiate y=x^x?
y=x^x
To find the differential, dy/dx, logs of both sides must be taken:
log(y)=log(x^x)
Then using log rules, the power can be brought down, outside the log expression:
log(y) = x log(x)
This expressions can now be differentiated with respect to x, using the chain rule on the left and the product rule on the right, giving:
(1/y) * dy/dx = 1 + log(x)
Multiplying through by y gives:
dy/dx = y (1 + log(x))
Remember! From the start of the question y=x^x, so this can be rewritten to:
dy/dx = x^x + x^xlog(x)
PL
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All answers ▸What exactly IS differentiation?
Differentiate 2x^3+23x^2+3x+5 and find the values of x for which the function f(x) is at either at a maximum or minimum point. (Don't need to specify which is which)
A curve is defined by the parametric equations: X = 3 – 4t , y = 1 + (2/t) Find (dy/dx) in terms of t.
Use integration by parts to find ∫ (x^2)sin(x) dx. (A good example of having to use the by parts formula twice.)