The Logarithmic Function: Calculus help?

I really don't know what you need for number 1 so I'm going to skip to questions 2 and 3.

2.) y = x^3*ln(x)

The slope at any point (x,y) on the curve can be written as:

dy/dx = x^2 + 3x^2*ln(x) (Product rule)

Slope of tangent line at (1, 0):

dy/dx eval. at x = 1 is 1.

Eqn of tangent line:

y = x - 1

3.) y = √(x^2 - 1)/√(x^2 + 1)

Taking the natural logarithm of both sides:

ln(y) = 1/2*ln(x^2 - 1) - 1/2*ln(x^2 + 1)

Differentiating both sides:

y'/y = x/(x^2 - 1) - x/(x^2 + 1)

y' = [√(x^2 - 1)/√(x^2 + 1)] * [ x/(x^2 - 1) - x/(x^2 + 1)]

y' = x/√(x^4 - 1) - x√(x^2 - 1)/(x^2 + 1)^(3/2)

(1) I am not sure what you want us to do with the function. Do you want us to differentiate it? Find the tangent line? What? Can you clarify on what you want us to do?

(2) Note that we already know a point on the line, which is (1, 0). In order for us to obtain the equation of the tangent line, we need to get its slope. To get the slope, we need to evaluate the derivative at x = 1.

By the Product Rule, we see that:

dy/dx = (x^3)'ln(x) + x^3[ln(x)]'

= 3x^2*ln(x) + x^3(1/x)

= 3x^2*ln(x) + x^2

= x^2[3ln(x) + 1].

At x = 1:

dy/dx (evaluated at x = 1) = 1(0 + 1) = 1.

Thus, the slope of the tangent line is 1. Therefore, using point-slope form, the equation of the tangent line is:

y - 0 = 1(x - 1) ==> y = x - 1.

(3) Taking the natural logarithm of both sides gives:

ln(y) = ln[√(x^2 - 1)/√(x^2 + 1)]

= ln[√(x^2 - 1)] - ln[√(x^2 + 1)], since ln(a/b) = ln(a) - ln(b)

= (1/2)ln(x^2 - 1) - (1/2)ln(x^2 + 1), since ln(a^b) = b*ln(a).

Then, differentiating both sides implicitly with respect to x gives:

(1/y)(dy/dx) = x/(x^2 - 1) - x/(x^2 + 1)

==> dy/dx = y[x/(x^2 - 1) - x/(x^2 + 1)].

Since y = √(x^2 - 1)/√(x^2 + 1):

dy/dx = √(x^2 - 1)/√(x^2 + 1) * [x/(x^2 - 1) - x/(x^2 + 1)].

This can be simplified if needed.

I hope this helps!