Derivative of Exponential and Logarithmic Functions

[domyy]

Homework Statement

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:

ye^x+ y² = 7lny - x³

Solution:

(e^x)(y') + 2y(y') - 7(y'/y) = -ye - 3x²

y'(e^x + 2y - 7/y) = -ye^x - 3x²
y' = -ye^x - 3x²/(e^x + 2y - 7/y)

 

[Ray Vickson]

Homework Statement

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:

ye^x+ y² = 7lny - x³

Solution:

(e^x)(y') + 2y(y') - 7(y'/y) = -ye - 3x²

y'(e^x + 2y - 7/y) = -ye^x - 3x²
y' = -ye^x - 3x²/(e^x + 2y - 7/y)

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

 

[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) isn't = x^2[1/x . (x)'] ?

 

[domyy]

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

 

[SammyS]

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) isn't = x^2[1/x . (x)'] ?

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

Use the product rule.

 

[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

 

[domyy]

Thanks a lot!

 

[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)

Instead of:

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?

 

[SammyS]

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)

Instead of:

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?

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(u^r) as (r)ln(u) reduces the number of times you need to use the chain rule.