Logarithmic Derivative of y = 5ln(7lnx)

[Jan Hill]

Homework Statement

y = 5ln(7lnx)

Homework Equations

y' = 5 x 1/7lnx +97lnx0 x 1/x

= 5(7lnx)/7lnx + 5ln x 7/x

The Attempt at a Solution

y' = 5 + 35ln/x

is this right?

 

[Mark44]

Homework Statement

y = 5ln(7lnx)

Homework Equations

y' = 5 x 1/7lnx +97lnx0 x 1/x

= 5(7lnx)/7lnx + 5ln x 7/x

You're completely forgetting the chain rule.
Where did the 97 come from?
What does lnx0 mean?

y' = 5 * 1/(7 lnx) * d/dx(7 lnx) = ?

The Attempt at a Solution

y' = 5 + 35ln/x

is this right?

 

[Jan Hill]

then I guess it should be

y' = 5/(7lnx) x 7/x

Y' = 35/x(7lnx)

Is that right?

 

[Mark44]

then I guess it should be

y' = 5/(7lnx) x 7/x

Y' = 35/x(7lnx)

Is that right?

Yes, but it should be simplified to 5/(x ln(x))