How do you find this second derivative?

Question

f(x)= (7e^x)/(7e^x+6)

f''(x) = ?

I'm able to derive the 1st derivative using the quotient rule, coming out to (42e^x)/((7e^x+6)^2)

but am unable to get the 2nd derivative using the quotient rule because it gets super messy and complicated.

Are there any easy tricks or patterns than can be used to derive a clean 2nd derivative? Because I will need to solve for inflection points.

Answer 1

Write f(x) = 7e^x/(7e^x + 6).

When taking the derivative of a function that is the ratio of two "sub-functions", the rule is "bottom sub-function times derivative of top sub-function minus top sub-function times derivative of bottom sub-function, all divided by bottom sub-function squared."

A popular memory aid for this rule is "low-dee-high minus high-dee-low, all over low-squared".

Next, f'(x) is [(7e^x + 6)7e^x − 7e^x(7e^x)]/(7e^x + 6)² or 42e^x/(7e^x + 6)².

Then f''(x) is [(7e^x + 6)²(42e^x) − (42e^x)(2)(7e^x)(7e^x + 6)/(7e^x + 6)⁴.

Extract factors of (7e^x + 6) from numerator and denominator to obtain f''(x) equal to

42e^x[(7e^x + 6) − 14e^x]/(7e^x + 6)³ or 42e^x[6 − 7e^x]/(7e^x + 6)³ or [252e^x − 294e^2x]/(7e^x + 6)³.

Answer 2

{(7x^x+6)^2 * (42e^x) - (42e^x) 2( 7e^x+6)} ( (7e^x) / (7e^x+6)^4 =

{(7x^x+6)(42e^x) - 2(42e^x)} / (7e^x+6)^3 =

(42e^x)( 7x^2+4) / (7e^x+6)^3

Answer 3

Use the product rule vs quotient rule by changing the first derivative to be 42e^x *(7e^x+6)^-2 ... The 2nd will then be 42e^2*(7e^x+6)^-2 -2*42e^x*(7)*(7e^X +6)^-3 ...