Difference quotient...

[k9fireman]

Anyone can help here????

find [f(x+h) - f(x)]/h if f(x)= x+2ln(x)

I get to

[(x+h) + 2ln(x+h) - (x+2ln(x))]/h

[(x+h) + 2ln(x+h) - x - 2ln(x)]/h help.....????

 

[galactus]

\(\displaystyle \L\\\frac{x+h+2ln(x+h)-(x+2ln(x))}{h}\)

\(\displaystyle \L\\=\frac{h+2ln(x+h)-2ln(x)}{h}\)

You could use L'Hopital's rule:

The derivative of the numerator with respect to h:

\(\displaystyle \L\\\frac{2}{x+h}+1\)

Of course, the derivative of the denominator wrt h is 1.

Now, \(\displaystyle \L\\\lim_{h\to\infty}(\frac{2}{x+h}+1)\)

 

[stapel]

[(x+h) + 2ln(x+h) - x - 2ln(x)]/h

Now keep going:

. . . . .[x + h + 2ln(x + h) - x - 2ln(x)] / h

. . . . .[x - x + h + 2(ln(x + h) - ln(x))] / h

Simplify the x's. Apply a log rule to the logs.

Do you have an answer (from the back of the book, maybe) that you're aiming for?

Thank you.

Eliz.

 

[k9fireman]

The answer is from a test i took for an online class, just didn't know how they got there. They had...

1+(2/h)ln[(x+h)/x]

Once i saw the ln(x+h) - ln(x) and got the ln[(x+h)/x] term, the h/h and 2/h became apparent. Thanks!!