How do I find the derivative of 7x (lnx +ln7)-8x+e?

Question

F (x)= 7x (lnx +ln7)-8x+e

I assumed e was a constant so it's derivative is 0, and the derivative of an ln constant is 0 so I got 

7 (lnx+ln7)*(1/x) -8

but I got it wrong and I'm a little stupped on how to approach this in a different manner. 

 

Answer 1

lnx + ln7 = ln(7x)

 

So, F(x) = 7xln(7x) - 8x + e

 

     F'(x) = 7ln(7x) + 7x(7/7x) - 8 + 0

 

            = 7ln(7x) + 7 - 8

 

            = 7ln(7x) - 1

Answer 2

Let us focus on the 7x(lnx + ln7) part.  Use the product rule and the ln rule.  

 

u = 7x

v = lnx + ln7

(uv)' = u'v + uv'

 

(uv)' = 7(lnx + ln7) + (7x)(1/x + 0) = 7(lnx + ln7) + 7