# Thread: In desperate need of some help!!

1. ## In desperate need of some help!!

Hey everyone...

I am about to start university next week, where I will have a maths test just testing our mathematical ability. However, since I have been on a gap year, my maths skills have gone somewhat rusty, and this is me being polite to myself!

So, I was wandering if you guys could help me with a few maths problems? Please only feel free to answer which ever ones you can, and I am terribly sorry I am asking such a huge favour without contributing anything first. I have got the answers to the questions, but just don't know how to arrive to them!

1) Rearrange the following formula to give d in terms of D and f:

(2 / D - d) + (2 / D + d) = (1/f)

It does look simpler if you write it out!
The answer is d = (D^2 - 4Df)^(1/2)

2) Given that y = loge (Ae^(-Bx)), find x in terms of y, A and B.

The answer is 1/B(loge A - y)

Differentiation Questions

3) Differentiate y = ln(3x^(2) +1) with respect to x.

4) Differentiate y = (x^(2) + x)^(-2) with respect to x

5) By differentiating the relation x^2 + y^2 = e^k (where k is a constant) with respect to x, find dy/dx.

I would really appreciate it you guys could help me out here, as I genuinely require some help.

2. Originally Posted by Singh_87
1) Rearrange the following formula to give d in terms of D and f:

(2 / D - d) + (2 / D + d) = (1/f)

It does look simpler if you write it out!
The answer is d = (D^2 - 4Df)^(1/2)
$\frac{2}{D - d} + \frac{2}{D + d} = \frac{1}{f}$

Multiply both sides by $f(D - d)(D + d)$.

$2f(D + d) + 2f(D - d) = (D - d)(D + d)$

$2fD + 2fd + 2fD - 2fd = D^2 - d^2$

$4fD = D^2 - d^2$

$d^2 = D^2 - 4fD$

$d = \sqrt{D^2 - 4fD}$

-Dan

3. 1) Rearrange the following formula to give d in terms of D and f:

(2 / D - d) + (2 / D + d) = (1/f)

It does look simpler if you write it out!
The answer is d = (D^2 - 4Df)^(1/2)

The idea is to do anything so that d will be isolated by itselt in the Lefthand side (LHS).

I think that should be
[2/(D-d)] +[2/(D+d)] = 1/f
Combine the two fractions in the LHS.
[2(D+d) +2(D-d)] /[(D-d)(D+d)] = 1/f
[2D +2d +2D -2d] /(D^2 -d^2) = 1/f
[4D] /(D^2 -d^2) = 1/f
Cross multiply,
(4D)(f) = (D^2 -d^2)(1)
4Df = D^2 -d^2
d^2 = D^2 -4Df
Take the square roots of both sides,
d = sqrt[D^2 -4Df]
Or,
d = [D^2 -4Df]^(1/2)

2) Given that y = loge (Ae^(-Bx)), find x in terms of y, A and B.

The answer is 1/B(loge A - y)

y = ln[A*e^(-Bx)]
y = ln[A] +ln[e^(-Bx)]
y = ln[A] +(-Bx)ln[e]
y = ln[A] -(Bx)(1)
y = ln[A] -Bx
Bx = ln[A] -y
x = (1/B)(ln[A] -y)

We used:
---Log(ab) = Log(a) +Log(b)
---Log(a^b) = b*Log(a)
---Log(to the base a)[a] = 1

Differentiation Questions

3) Differentiate y = ln(3x^(2) +1) with respect to x.

y = ln[3x^2 +1]
dy/dx = (1/[3x^2 +1])*(6x)
dy/dx = (6x) /(3x^2 +1)

4) Differentiate y = (x^(2) + x)^(-2) with respect to x

y = (x^2 +x)^(-2)
dy/dx = {(-2)[x^2 +x]^(-3)}*(2x +1)
dy/dx = (-4x -2)[x^2 +x]^(-3)
dy/dx = -(4x +2)[x^2 +x]^(-3)

5) By differentiating the relation x^2 + y^2 = e^k (where k is a constant) with respect to x, find dy/dx.

x^2 +y^2 = e^k
Differentiate both sides,
(2x)dx +(2y)dy = 0
(x)dx +(y)dy = 0
Divide both sides by dx,
x +y(dy/dx) = 0
y(dy/dx) = -x
dy/dx = -x/y

4. Originally Posted by Singh_87
2) Given that y = loge (Ae^(-Bx)), find x in terms of y, A and B.

The answer is 1/B(loge A - y)
The standard symbol (as I learned it, anyway) for $log_e$ is $ln$. Thus
$y = ln(Ae^{-Bx})$ <-- $ln(ab) = ln(a) + ln(b)$

$y = ln(A) + ln(e^{-Bx})$ <-- $ln(a^b) = b \cdot ln(a)$

$y = ln(A) + -Bx \cdot ln(e)$ <-- $ln(e) = 1$

$y = ln(A) - Bx$

$Bx = y - ln(A)$

$x = \frac{y - ln(A)}{B}$

-Dan

5. Originally Posted by Singh_87
3) Differentiate y = ln(3x^(2) +1) with respect to x.

$y = ln(3x^2 + 1)$

Recall that $\frac{d}{dx}ln(x) = \frac{1}{x}$. So applying the chain rule:
$\frac{dy}{dx} = \frac{1}{3x^2 + 1} \cdot (6x) = \frac{6x}{3x^2 + 1}$

-Dan

6. Originally Posted by Singh_87
4) Differentiate y = (x^(2) + x)^(-2) with respect to x

Again, use the chain rule:
$y = (x^2 + x)^{-2}$

$\frac{dy}{dx} = (-2) \cdot (x^2 + x)^{-3} \cdot (2x + 1) = -(4x + 2)(x^2 + x)^{-3}$

-Dan

7. Originally Posted by Singh_87
5) By differentiating the relation x^2 + y^2 = e^k (where k is a constant) with respect to x, find dy/dx.

This is an implicit differentiation problem. Note that when we use the chain rule to find the derivative of a function $f(y)$ with respect to x we get $\frac{df}{dy} \cdot \frac{dy}{dx}$, so
$x^2 + y^2 = e^k$ <-- Notice that the right hand side is merely a constant!

Taking the derivative of both sides gives:
$2x + 2y \cdot \frac{dy}{dx} = 0$

$2y \frac{dy}{dx} = -2x$

$\frac{dy}{dx} = \frac{-2x}{2y}$

$\frac{dy}{dx} = - \frac{x}{y}$

-Dan