Question

solve this problem:

Answer 1

well you will need to use chain rule, and quient rule

Answer 2

Just break derivatives up into tiny functions and them plug them into the formula

we know that

ln(x) = x'/x

Answer 3

and the derivative of Cf(x) is just Cf'(x)

Answer 4

so dont worry about that 1/3

Answer 5

where C is just a constant

Answer 6

so first thing you should do is write out what you know based on the derivative for ln(x)


\[f'(x) = \frac{x'}{\frac{x^2-x-20}{(x+7)^4}}\]

Answer 7

x' is probably not the right variable to use but oh well

Answer 8

okay but how did you find x' ?

Answer 9

First thing I do when I deal with a derivative is I split it into smaller functions and take the derivative of those smaller functions

So firstly we have the function

f(x) = (1/3)ln(x^2 - x - 20/2x - 1)
so we have to split it into different functions and use chain rule, outside l(x) inside o(x)

l(x) = 1/3ln(x)
l'(x) = (1/3)(x'/x)
so by chain rule we have

f'(x) = (1/3)*(o'(x)/o(x))
whats inside is really messy lets call it o(x)

o(x) = (x^2 - x - 20)/(2x - 1)

I'm going to split this into two more functions

g(x) = x^2 - x - 20
g'(x) = 2x - 1

s(x) = (x+7)^(4)
To find s'(x) we need to use CHAIN RULE so again I split it into smaller functions

d(x) = x^(4)
d'(x) 4x^3

m(x) = x + 7
m'(x) = 1

we know chain rule is just d'(m(x)) * m'(x) = s'(x)
so
s'(x) = 4(x+7)^(3)*1

so now we have

g(x) = x^2 - x - 20
g'(x) = 2x - 1

s(x) = (x+7)^(4)
s'(x) = 4(x+7)^(3)

ok now we just need to apply quient rule

which is

o'(x) = ( g'(x)s(x) - g(x)s'(x) )/(s(x))^2

sub in the values

o'(x) = ( (2x - 1)(x+7)^(4) - (x^2 - x - 20)4(x+7)^(3) )/( (x+7)^(4) )^2

now
go back to the original stuff we figured out

l(x) = 1/3ln(x)
l'(x) = (1/3)*(x'/x)

o(x) = (x^2 - x - 20)/(2x - 1)
o'(x) = ( (2x - 1)(x+7)^(4) - (x^2 - x - 20)4(x+7)^(3) )/( (x+7)^(4) )^2

plug it all into f'(x) = (1/3)*(o'(x)/o(x))

Answer 10

this is the long way to do derivatives but if you do it this way you will never have a problem with them again

Answer 11

also after sometime you wont need to write it out you will just be able to do it in your head

Answer 12

oh thank you so much for your effort :)

Answer 13

do you follow?

Answer 14

if you get confused please feel free to ask a question

Answer 15

sure :) thanks again