that's the way it is usually done.
lny = 1/x (1/2 lnx)
1/y y' = -1/x^2 (1/2 lnx) + 1/x (1/2x)
= -1/(2x^2) (lnx + 1)
y' = -sqrt(x)^(1/x) / (2x^2) (lnx+1)
or
-1/2 x^(1/(2x)-2) (lnx+1)
Using the same method, you can work out that if
y = u^v
y' = v u^(v-1) u' + lnu u^v v'
You can see that it a combination/generalization of the derivatives of
u^n and a^u for constants a and n.
y=sqrt(x)^1/x. Find dy/dx
That is to the power of 1/x
So to start this do I need to take Ln on both sides and then use a logarithm property?
Ln(y)=lnx^1/2x?
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