# Derivative of Exponential and Logarithmic Functions

1. Jan 20, 2013

### domyy

1. The problem statement, all variables and given/known data

PROBLEM 1
Find slope of tangent line at (1,0)

g(x) = x2lnx

Solution:

g'(x) = (x2lnx)'
g'(x) = x2[1/x . (x)']
g'(x) = x2 . 1/x
g'(x) = x

M tan/(1,0) = 0

PROBLEM 2

Use implicit differentiation:

yex+ y2 = 7lny - x3

Solution:

(ex)(y') + 2y(y') - 7(y'/y) = -ye - 3x2

y'(ex + 2y - 7/y) = -yex - 3x2
y' = -yex - 3x2/(ex + 2y - 7/y)

Last edited: Jan 20, 2013
2. Jan 20, 2013

### Ray Vickson

Your derivative of $x^2 \ln(x)$ is wrong.

3. Jan 20, 2013

### domyy

Thanks for the reply. That's why posted both problems. I want to understand why.
If the derivative of 7lny on the second problem is 7[(1/y . (y)']
then why x^2 ln(x) isnt = x^2[1/x . (x)'] ?

4. Jan 20, 2013

### domyy

OHHHH I THINK I GOT IT! Am I supposed to use the product rule?

5. Jan 20, 2013

### SammyS

Staff Emeritus
The derivative of 7 is zero, because 7 is a constant . x2 is not a constant.

Use the product rule.

6. Jan 20, 2013

### domyy

Ok, if so...this is how I'd do it:

g(x) = x2lnx

g'(x) = (x2)(1/x) + (lnx)(2x)
g'(x) = x + 2xlnx
g'(x) = 1 + 2(1)ln(1)
g'(x) = x + 2ln1
g'(x) = 1 + 0 =1

7. Jan 20, 2013

### domyy

Thanks a lot!

Last edited: Jan 20, 2013
8. Jan 20, 2013

### domyy

I have another question:

Is it wrong to solve the problem this way:

y = ln (√x3- 5)

y' = [(x3 - 5)1/2]'/[x3 - 5]

y' = 1/2(x3 - 5)-1/2 (x3 - 5)'/ (x3 - 5)

y' = 3x2/ 2[(√x3 - 5)(√x3 - 5)]

y' = 3x2/ 2(x3 - 5)

y = ln (x3 - 5)1/2 = 1/2 ln (x3 - 5)
y' = 1/2 . (x3 - 5)'/(x3 - 5)
y' = 3x2/2(x3 - 5)

Can I solve it either way or could the first method cause me further problems with other exercises?

Last edited: Jan 20, 2013
9. Jan 20, 2013

### SammyS

Staff Emeritus
It's best to start a new thread for a new problem.

Also, use sufficient parentheses.
√(x3-5) ≠ √x3-5 .​

As for your question: Either method is fine, but writing ln(ur) as (r)ln(u) reduces the number of times you need to use the chain rule.

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