# Calculus Help: ln(x) differentiating program

Hi Tom,

I found a problem with your ln(x) differentiating program. It always solves g'(x) by saying g'(x)=u’/u, right? But even if u’/u contains a variable, the program will solve u’/u as an integer.

Here are some examples:

- If , find

I calculated using program X:ln(x), option 1: Differentiating

The program uses the Product Rule, and eventually solves for g'(x)

It says g'(x)=u’/u

u’=7

u=7x

g'(x)=2

When g'(x) is actually =7/7x, or 1/x

2. If , find

I calculated using program X:ln(x), option 1: Differentiating

The program uses the Product Rule, and eventually solves for g'(x)

It says g'(x)=u’/u

u’=1

u=x+6

g'(x)=0

When g'(x) is actually =1/x+6

3. If , find

I calculated using program X:ln(x), option 1: Differentiating

The program uses the Product Rule, and eventually solves for g'(x)

It says g'(x)=u’/u

u’=1

u=x

g'(x)=2

When g'(x) is actually =1/x

As you can imagine, when g'(x) is wrong,the whole answer is wrong too. Anyhow, fyi.

Thanks,

Sarah

Sarah-

As far as problem 1,

It solves for h ‘ (x) because f(x) and g(x) are in the equation

f(x) = 3

f ‘ (x) = 0

g(x) = ln(7x)

g ‘ (x) = u’ / u = 7/7x = 1/x

Everything seems ok

Calculator gets the same answer

h ‘ (x) = 3/x

Problem 2

There’s always some trick the professors don’t emphasize, so I learned something again

evidently when there is a plus sign or minus sign within the ln ( ) you switch f(x) and the g(x). I did this in my program to update – give yourself a hug

answer

5/(x+1)

Problem 3

Seems to be ok

same answer as calclulator

Tom

## Leave a Reply