Could I please get some help for these 3 questions, apologies for asking so many, but they were the ones I was really unsure of:

1. Find an equation for the line tangent to y=ln(x) and that goes through the origin.

I have no idea, the ln(x) stuff is confusing me.

2. derivative of y=log_3(1+xln(3))? (the _3 means base 3). I got 1/(ln(3)*(ln(3)+(x/3))/(1+xln(3)), but I don't think i did this right- is the second part of the product rule where we take the derivative of the ln(3) multiply by x supposed to be 0 or x/3? Could you show your work to me too to check?

3. Derivative of ln(x/(√ (x+1)))? I don't know how to simplify further: (√ (x+1))/x * (√ (x+1))/(x+1) - x/[(2(x+1)(√ (x+1))].

1.

y = lnx
dy/dx = 1/x

let the point of contact be (a, lna)
at (a, lna) , the slope is 1/a

slope using the grade 9 way of doing slope
= (lna - 0)/(a-0) = lna/a

but lna/a = 1/a
lna = 1
so a = e , (which is appr 2.71828..)

so we want the equation of a line with slope 1/e and point (e, 1)
y-1 = (1/e)(x-e)
fix up any way you need to

2. y = log3 (1 + xln3)

dy/dx = (1/ln3) (ln3/(1+xln3)
= 1/(1 + xln3)

3. May I suggest you use rules of logs and change
y = ln (x/√(x+1) ) to
y = lnx - ln √(x+1)
= lnx - (1/2)ln(x+1)

now take the derivative

No problem! I'm here to help you with your questions. I'll break down each question and explain how to solve it step by step.

1. Find an equation for the line tangent to y = ln(x) that goes through the origin:
To find the equation of a tangent line, we need to find the slope of the tangent and a point on the line (in this case, the origin).

Step 1: Find the derivative of the function y = ln(x).
The derivative of ln(x) is 1/x. So, the slope of the tangent line at any given point is 1/x.

Step 2: Find the equation of the tangent line.
Since the line goes through the origin (0,0), we have a point (0,0) and the slope 1/x. The equation of a line can be represented as y = mx + b, where m is the slope and b is the y-intercept.

Substituting the values, we get:
0 = (1/x)*0 + b
0 = b
So the equation of the line tangent to y = ln(x) that goes through the origin is y = 1/x.

2. Derivative of y = log_3(1 + xln(3)):
To differentiate this function, we'll use the chain rule along with the logarithmic differentiation.

Step 1: Determine the derivative of the outer function.
The derivative of log_3(u) with respect to u is 1/(u * ln(3)), where u is the argument of the logarithm.

Step 2: Apply the chain rule.
Let u = 1 + xln(3).
Now, we can differentiate the function using the chain rule.

dy/dx = (1/(u * ln(3))) * (d/dx(u))
= (1/(u * ln(3))) * (1ln(3) * x + 1 * ln(3))
= (1/(u * ln(3))) * (ln(3)x + ln(3))
= (ln(3)x + ln(3))/(u * ln(3))
= (ln(3)x + ln(3))/(1 + xln(3))

Your initial result is correct. The second part of the product rule should be x/3.

3. Derivative of y = ln(x/√(x + 1)):
To simplify this expression, let's first rewrite it in a simpler form.

Step 1: Simplify the expression.
Consider that √(x + 1) is the same as (x + 1)^(1/2).
So, y = ln(x/(x + 1)^(1/2))
= ln[(x/((x + 1)^(1/2)))]

Step 2: Apply the logarithmic differentiation.
Using the chain rule and logarithmic differentiation, we'll differentiate the function.

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

Therefore, the derivative of y = ln(x/√(x + 1)) is 1.

I hope this helps! Let me know if you have any further questions.