how do I differentiate log_base2 ((x^2e^2/2sqrt(x+1))?

Question

If anyone can explain briefly on each step, I'd very appreciate it. thank you

Answer 1

At first I thought it would be really ugly, but...I think it's not too awful really if you break it into pieces first using laws of logs.

log2(x²e²/(2√(x + 1))) =

log2(x²e²) - log2(2√(x + 1)) =

log2(x²) + log2(e²) - [log2(2) + log2(√(x + 1))] =

2 log2(x) + 2 log2(e) - [1 + 1/2 log2(x + 1))] =

2 log2(x) + 2 log2(e) - 1 - 1/2 log2(x + 1)

And then, according to this site (http://people.hofstra.edu/faculty/Stefan_Waner/realworld/tutorials/unit3_3.html ), the derivative of log2(x) is 1/[x ln(2)], so..

[2 log2(x) + 2 log2(e) - 1 - 1/2 log2(x + 1)]' =

2/[x ln(2)] + 0 - 0 - 1/[2(x + 1)ln(2)] =

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

1/ln(2)[(4(x + 1) - x)/(2x(x + 1)] =

(3x + 4)/[ln(2)(2x² + 2)], or

(3x + 4)/[ln(4)(x² + 1)], as kaksi posted.

Answer 2

♠ y(x)=log_base2 ((x^2*e^2 /(2sqrt(x+1))) = log_base2 (f(x));
let us get rid off base2 and translate to base_e: 2^y =f(x) and y*ln2 =ln(f(x)) and
y(x) = (1/ln2)*{ln(x^2) +ln(e^2) –ln2 –ln(sqrt(x+1))} =
=(1/ln2)*{2lnx +2 –ln2 –0.5ln(x+1)}, hence
♣ y’=(1/ln2)*(2/x –0.5/(x+1)) = (1/2ln2)*(4x+4 –x)/(x*(x+1)) =
= (1/ln4)* (3x+4) /(x*(x+1)); friends? Beer on your part.