Logarithmic Derivative of y = 5ln(7lnx)

Jan Hill
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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?
 
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Jan Hill said:

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) = ?
Jan Hill said:

The Attempt at a Solution



y' = 5 + 35ln/x

is this right?
 
then I guess it should be

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

Y' = 35/x(7lnx)

Is that right?
 
Jan Hill said:
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))
 
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