Expert: Paul Klarreich - 2/22/2007

Question
I having so much trouble out of some of these problems can you help me please?
y=[ln(e to the 2x+1]raised to 2
y=ln2*ln x
f(x)= 4x squared over x squared + e to the 2x
Thank you

Answer
Questioner:   Larr
Category:  Calculus
 
Subject:  derivatives of logarithmic and exponential functions

>> Good subject line.

Question:  I having so much trouble out of some of these problems can you help me please?
y=[ln(e to the 2x+1]raised to 2
y=ln2*ln x
f(x)= 4x squared over x squared + e to the 2x
Thank you
...............................
Hi, Larr,

Don't forget to include instructions with the examples.  I assume, from the subject line, that in each case your job is to find the derivative.

y = [ ln(e^(2x+1)) ]^2.

Now a basic property of these functions is that   ln (e^u) = u.  Also, e^(ln u) = u.

So this example becomes:

y = (2x + 1)^2

I think you can take it from there.
............................

y = ln 2  ln x

ln 2 is just a constant, approximately  0.69

y = 0.69 ln x

Assuming you know how to differentiate  ln x, the rest is easy.

.......................

f(x)= 4x squared over x squared + e to the 2x
        4x^2
f(x) = ----------
      x^2 + e^2x

This is a quotient rule example.
       (x^2 + e^2x)(8x) - (4x^2)(2x + 2e^2x)
f'(x) = --------------------------------------
              (x^2 + e^2x)^2

       (x^2 + e^2x)(8x) - (8x^2)(x + e^2x)
f'(x) = --------------------------------------
              (x^2 + e^2x)^2

       8x[(x^2 + e^2x) - (x)(x + e^2x)]
f'(x) = --------------------------------------
              (x^2 + e^2x)^2

       8x[x^2 + e^2x - x^2 - xe^2x]
f'(x) = ------------------------------
              (x^2 + e^2x)^2

       8x[e^2x - xe^2x]
f'(x) = ------------------
        (x^2 + e^2x)^2

       8x e^2x[1 - x]
f'(x) = ---------------
       (x^2 + e^2x)^2
That's about it.