What are the derivatives of the following Problems?

Question

1. Logb20 X
- B stands for base , so it reads Log with base 20 (x).
2.Y = ln (x^3+2x^2+4)
3.Y = ln[1+e^(x)/1-e^(x)]
4.Y = e^(x^(2+1)) * ln X (most difficult)

10 points for Best Helper

Yes, It does appear that I did make a mistake on #4. It was supposed to be
y = e^(x^(2)+1) * ln X, sorry.

Answer 1

1. log_20 x = ln x / ln 20
so d/dx (log_20 x) = (1/x) / ln 20 = 1 / (x ln 20)

2. y = ln (x^3 + 2x^2 + 4)
dy/dx = 1/(x^3 + 2x^2 + 4) . d/dx (x^3 + 2x^2 + 4) by the chain rule
= (3x^2 + 4x) / (x^3 + 2x^2 + 4)

3. y = ln [(1 + e^x) / (1 - e^x)]
dy/dx = (1 - e^x) / (1 + e^x) . d/dx [(1 + e^x) / (1 - e^x)]
= {(1 - e^x) / (1 + e^x)} . [e^x (1 - e^x) - (1 + e^x) (-e^x)] / (1 - e^x)^2
= {1 / (1 + e^x)} . [e^x - e^(2x) + e^x + e^(2x)] / (1 - e^x)
= 2e^x / [(1 + e^x) (1 - e^x)]
= 2e^x / (1 - e^(2x)).

I assume the bracketing in (4) is incorrect and should be:
4. y = e^(x^2 + 1) . ln x
Otherwise, x^(2+1) = x^3 so it would normally have been written simply as y = e^(x^3) . ln x.

y = e^(x^2 + 1) . ln x
so dy/dx = e^(x^2 + 1) . (1/x) + (ln x) e^(x^2+1) . (2x) by the product rule and chain rule.
= e^(x^2 + 1) [1/x + 2x ln x].

Answer 2

1. We can rewrite the expression using the change of base formula to be:

ln(x) / ln(20), which is equivalent to
(1/ln(20)) (ln(x)). And so our derivative is
(1/ln(20))(1/x), or simply 1 / (x*ln(20)).


2. If we use the chain rule, we get the derivative of the outer function (ln(x)) evaluated at the inner function (x^3 + 2x^2 + 4) times the derivative of the inner function. Since the derivative of ln(x) = 1/x, we get:

1/(x^3 + 2x^2 +4) * (3x^2 + 4x), or simply
(3x^2 + 4x) / (x^3 + 2x^2 + 4).

3. Using the chain rule, we get derivative of the outer (ln(x)) evaluated at the inner [1+ e^x / 1 - e^x] times the derivative of the inner (which, using quotient rule, is [(1-e^x)e^x - (1+e^x)-e^x] / (1-e^x)^2 [Remember, it's Lo-dee-Hi minus Hi-dee-Lo over Lo-squared where Hi is the numerator and Lo is the denominator. And dee of course is the derivative]. So combining, we get:

1/[1+e^(x)/1-e^(x)] * [(1-e^x)e^x - (1+e^x)-e^x] / (1-e^x)^2

We can reduce the first term to:
[1-e^(x) / 1+e^(x)] and the second to:
2e^x / (1-e^x)^2. Substituting, we get:

2e^x [1-e^(x)] / [(1+e^(x))(1-e^x)^2)] and finally

2e^x / [(1+e^x)(1-e^x)]

The last one is written wrong. If it were solved as written, the x^(2+1) would simplify to just x^3, which would make the problem WAAAAY easy. So check your typing on that one and get back in touch with me. :)