Math Is Fun Forum

fusilli_jerry89

Member

Registered: 2006-06-23

Posts: 86

Derivatives Again...Sorry uggh

y=e^(1+lnx)    e^(1+lnx) • x-¹    (e^(1+lnx))/x The answer in the back is just e. How do you simplify what i got further to e?

y=x^(lnx) i know the derivative of lnx is 1/x but what about x^(lnx)?

ln(cos-¹x)  (1/(cos-¹x))(-x/√(1-x²))       -x/(cos-¹x√(1-x²))
The back in the book has the same answer except the -x is a -1 instead.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Derivatives Again...Sorry uggh

You've started the first question off correctly, by multiplying by the derivative of the power of e.

However, as the answer book says, it can be simplified a bit.

e^(lnx + 1)/x = (e^lnx*e^1)/x = (x*e)/x = e.

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fusilli_jerry89

Member

Registered: 2006-06-23

Posts: 86

Re: Derivatives Again...Sorry uggh

any help on the others?

LQ

Real Member

Registered: 2006-12-04

Posts: 1,285

Re: Derivatives Again...Sorry uggh

Overall, e^(ln(x)+1)=e^(ln(x) + 0)*e^n(0+1). the 0 can also be dealt with as the 1, in fact any additive number n in the exponent means an added multiple q^n

As you can see, when this is given, all equations that means e^(ln(x)+1)/x is e.

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