Derivatives, must be easy, huh?

Question

Find f'(x), and no, don't just use your calculator, I can do that too.
Please show your work, because I can't seem to get this.

f(x) = ln ( 7x + (49x^2 - 16)^(1/2))

Answer 1

using the rule

df/dx = df/dy * dy/dx and y = 7x +(49x^2-16)^1/2

we get df/dy = 1/y = 1/(7x+49x^2-16)^1/2 ...1

dy/dx = 7+ 1/2(49x^2-16)^(-1/2) d/dx(49x^2-16)
= 7+ 1/(2*49x^2-16)^(1/2) * 98 x ...2

using d/dx(y^n) = n y^(n-1) dy/dx where y = 49x^2-16

from 1and 2 by multiplying
df/dx = (7+ 98x/(2*49x^2-16)^(1/2))/(7x+49x^2-16)^1/2 ...

Answer 2

f(x) = ln ( 7x + (49x^2 - 16)^(1/2))
f'(x) = 1/ ( 7x + (49x^2 - 16)^(1/2)) { 7 + 1/2 (49x^2 - 16)^(-1/2) 98x}
= { 7 + 49x (49x^2 - 16)^(-1/2) } / ( 7x + (49x^2 - 16)^(1/2))
now you just need to simplify '

Answer 3

I would use substitution, and the fact that the derivative of lnx = 1/x

Answer 4

dy/dx = 1/(7x + (49x^2-16)^0.5) [d/dx(7x + (49x^2-16)^0.5)]

7 + (.5(49x^2-16)^-0.5)98x
= -------------------------------------
7x + (49x^2-16)^0.5