Calculus Natural Logarithms

Question

Find f'(x) given f(x) = ln(√(2x+9)(9x-3)). Note*both functions are under the square root.

I believe the problem needs to be set up like this = ln(2x+9)^1/2(9x-3)^1/2 and then = ln(2x+9)^1/2 + ln(9x-3)^1/2

But I am not sure. Hopefully someone can help me out with this. Thanks

Answer 1

Find f'(x) given f(x) = ln(√(2x+9)(9x-3)).

Rewrite  f(x) = ln(√(2x+9)(9x-3)) as  f(x) = ln[(2x+9)(9x-3)]^(1/2) = (1/2) ln[(2x+9)(9x-3)]

                    = (1/2)[ ln(2x+9) + ln (9x-3)]

From this point, you can easily differentiate the two ln( ) functions.

 

Answer 2

Rewrite the function.

 

 

f(x) = ln[((2x + 9)(9x - 3))^1/2]

 

f(x) = ln[(18x² + 75x - 27)^1/2]

 

 

Use the chain rule to derive.

 

f'(x) = [(1/2)(18x² + 75x - 27)^-1/2 (36x + 75)] / (18x² + 75x - 27)^1/2

 

 

Now we simplify as much as possible.

 

f'(x) = (36x + 75) / [2(18x² + 75x - 27)]

 

f'(x) = [3(12x + 25)] / [2 * 3(6x² + 25x - 9)]

 

f'(x) = [3(12x + 25)] / [6(6x² + 25x - 9)]

 

f'(x) = (12x + 25) / [2(6x² + 25x - 9)]

 

f'(x) = (12x + 25) / (12x² + 50x - 18)