Calc Midterm, First Year University (derivative related stuff)

y2klimen
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Homework Statement


If f(x)=x^x for x>0, find the constant a such that f'(a)=2f(a)


Homework Equations





The Attempt at a Solution


f'(a)=a*a^(a-1)
=a^a

2f(a)=2a^a

2a^a=a^a
ln2a^a=lna^a
aln2a=alna

...?
 
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If f(x) = x^x, then f'(x) =/= x*x^(x-1). That rule works if your exponent is a number, not a function. Rewrite x^x in exponential form and then differentiate using the chain rule to find f'(x).
 
y2klimen said:

Homework Statement


If f(x)=x^x for x>0, find the constant a such that f'(a)=2f(a)


Homework Equations





The Attempt at a Solution


f'(a)=a*a^(a-1)
=a^a

2f(a)=2a^a

2a^a=a^a
ln2a^a=lna^a
aln2a=alna

...?
No that is not right. You cannot treat a (x) as a variable in one case (the base) and a constant in the other (the exponent).

If y= xx then ln(y)= x ln(x). Use the chain rule to find dy/dx.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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