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
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 answer
1 answer