- #1

- 1

- 0

1. (a) Differentiate y = (x - A)/(x - B)

(b) Show that for A>B, all tangents have a positive gradient, and for A<B all tangent have a negative gradient

(c) What happens when A = B

**My attempt:**

(a) y' = (x-b)(1) - (x-A)(1) / (x - B)^2

y' = (2x - AB) / (x - B)^2

(a) y' = (x-b)(1) - (x-A)(1) / (x - B)^2

y' = (2x - AB) / (x - B)^2

2. Evaluate f'(1) when

(a) f(x) = (√x) + (√2) / (√x) - (√2)

(b) f(x) = (2x-3)(√x + 1)/ x

**My attempt:**

(a) f'(x) = (√x-√2)(1/2√x)-(√x+√2)(1/2√x)/(√x -√2)^2

= 1/2 - √2/2√x - 1/2 + √2/2√x

= (-√2/√x)/ (√x-√2)^2

f'(1) = (-√1) / (√1-√2)^2

(-√1) = (√1-√2)^2

I'm not sure what to do from here

(b) f(x) = (2x-3)(√x + 1)/ x

u = (2x-3)(√x + 1)

v = x

f'(x) =

I don't know how to find the derivate to

(a) f'(x) = (√x-√2)(1/2√x)-(√x+√2)(1/2√x)/(√x -√2)^2

= 1/2 - √2/2√x - 1/2 + √2/2√x

= (-√2/√x)/ (√x-√2)^2

f'(1) = (-√1) / (√1-√2)^2

(-√1) = (√1-√2)^2

I'm not sure what to do from here

(b) f(x) = (2x-3)(√x + 1)/ x

u = (2x-3)(√x + 1)

v = x

f'(x) =

I don't know how to find the derivate to

__u__. If I did I would only be able to get up to what I did in question 2. (a)With Q1 (b) and (c) am I supposed to equate the equation to A or B, like in quadratic equations? Or is that what I am supposed to do in Question 2. I am confused by what these questions are asking. Sorry if I haven't shown as much working out as I should have.

Thank you for reading