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 ![f(x) = 5 \ln(6+x)]()
, find ![f'( x )]()
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
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