• Jan 28th 2010, 07:32 AM
jimgreenmore
Differentiate the following
1. (i)$\displaystyle \frac{3}{x^2}$
I struggled with these because I'm used to differentiating square roots and fractions.

(ii) $\displaystyle y = x+\sqrt{x}$

2. Find the gradient of the curve $\displaystyle y = x^2 + \frac{1}{x}$ at the point (1,2)

3. Differentiate
(i) $\displaystyle y=(x+2)(x^3+5x)$
Answer I got = $\displaystyle 5x^3 + 10x - 6x^2 +10$

(ii) $\displaystyle y=\frac{3x+2}{2x+1}$
Answer I got = $\displaystyle \frac{7}{(2x+1)^2}$

Are my answers for the ones completed correct?
• Jan 28th 2010, 11:36 AM
earboth
Quote:

Originally Posted by jimgreenmore
Differentiate the following
1. (i)$\displaystyle \frac{3}{x^2}$
I struggled with these because I'm used to differentiating square roots and fractions.

$\displaystyle \frac3{x^2} = 3 x^{-2}$

Now use the chain-rule.
Quote:

(ii) $\displaystyle y = x+\sqrt{x}$

2. Find the gradient of the curve $\displaystyle y = x^2 + \frac{1}{x}$ at the point (1,2)
How did you get this result. Btw I've got 1.
Quote:

3. Differentiate
(i) $\displaystyle y=(x+2)(x^3+5x)$
Answer I got = $\displaystyle 5x^3 + 10x - 6x^2 +10$
If I were you I would have expanded the brackets before differentiation (otherwise you have to use the product rule and the chain-rule)

$\displaystyle y=(x+2)(x^3+5x) = x^4+2x^3+5x^2+5x$
Quote:

(ii) $\displaystyle y=\frac{3x+2}{2x+1}$
Answer I got = $\displaystyle \frac{7}{(2x+1)^2}$
You have to use the quotient rule (what you obviously attempted to do), but you must have made a sign mistake:

$\displaystyle y'=- \frac1{(2x+1)^2}$

Quote:

Are my answers for the ones completed correct?
No. See my remarks.
• Jan 28th 2010, 12:20 PM
jimgreenmore
Here is how I did them originally:

Differentiate the following

2. Find the gradient of the curve $\displaystyle y = x^2 + \frac{1}{x}$ at the point (1,2)

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

3. Differentiate
(i) $\displaystyle y=(x+2)(x^3+5x)$
$\displaystyle =(x+2)(3x^2+5)+(x^3+5x)(1)$
$\displaystyle =4x^3+5x-6x^2+10+x^3+5x$
$\displaystyle =5x^3 + 10x - 6x^2 +10$

(ii) $\displaystyle y=\frac{3x+2}{2x+1}$
$\displaystyle =\frac{dy}{dx} = 3x+2$
$\displaystyle =\frac{dy}{dx} = 3$

$\displaystyle =\frac{dy}{dx} = 2x+1$
$\displaystyle =\frac{dy}{dx} = 2$

$\displaystyle =\frac{(2x+1)3 - (3x+2)2}{(2x+1)^2}$
$\displaystyle =\frac{(6x+3-6x+4)}{(2x+1)^2}$

$\displaystyle =\frac{7}{(2x+1)^2}$

I did this after your help, $\displaystyle \frac{3}{x^2}$

I got:

$\displaystyle = 3x^{-2}$
$\displaystyle = -2(-3x)^{-3}(-3)$
$\displaystyle = 6(-3x)^{-3}$

Thanks for the help so far.
• Jan 28th 2010, 09:56 PM
earboth
Quote:

Originally Posted by jimgreenmore
Here is how I did them originally:

Differentiate the following

2. Find the gradient of the curve $\displaystyle y = x^2 + \frac{1}{x}$ at the point (1,2)

$\displaystyle \frac{dy}{dx} = 2x+\bold{\color{red}(-1)}x^{\bold{\color{red}-2}}$
$\displaystyle =2(1) + \bold{\color{red}(-1)}(1)^{\bold{\color{red}-2}}$
$\displaystyle =2\bold{\color{red}-1}$
$\displaystyle = \bold{\color{red}1}$

3. Differentiate
(i) $\displaystyle y=(x+2)(x^3+5x)$
$\displaystyle =(\bold{\color{red}x}+2)(\bold{\color{red}3x^2}+5) +(x^3+5x)(1)$ <<<<<< here you calculate $\displaystyle 1 \cdot 3 = 4$. Quite unusual :D
$\displaystyle =\bold{\color{red}4x^3}+5x-6x^2+10+x^3+5x$
$\displaystyle =5x^3 + 10x - 6x^2 +10$

(ii) $\displaystyle y=\frac{3x+2}{2x+1}$
$\displaystyle =\frac{dy}{dx} = 3x+2$
$\displaystyle =\frac{dy}{dx} = 3$

$\displaystyle =\frac{dy}{dx} = 2x+1$
$\displaystyle =\frac{dy}{dx} = 2$

$\displaystyle =\frac{(2x+1)3 - (3x+2)2}{(2x+1)^2}$
$\displaystyle =\frac{(6x+3-6x\bold{\color{red}-}4)}{(2x+1)^2}$

$\displaystyle =\frac{\bold{\color{red}-1}}{(2x+1)^2}$

I did this after your help, $\displaystyle \frac{3}{x^2}$

I got:

$\displaystyle = 3 \cdot x^{-2}$ <<<<<<<< The number 3 is a constant factor.
$\displaystyle = \bold{\color{red}3 \cdot (-2)}(x)^{-3}$
...

Thanks for the help so far.

...
• Feb 1st 2010, 02:08 AM
jimgreenmore
I'm still lost here. (Thinking)
• Feb 1st 2010, 02:43 AM
e^(i*pi)
Quote:

Originally Posted by jimgreenmore
(ii) $\displaystyle y = x+\sqrt{x}$ at the point (1,2)

As you know the power rule for differentiation is $\displaystyle \frac{d}{dx}x^n = nx^{n-1}$. From the laws of exponents $\displaystyle \sqrt{x} = x^{1/2}$

Therefore you get $\displaystyle f'(x) = 1 + \frac{1}{2}x^{(1/2) - 1}$

$\displaystyle f'(1) = 1+\frac{1}{2\sqrt1} = \frac{3}{2}$

Quote:

3. Differentiate
(ii) $\displaystyle y=\frac{3x+2}{2x+1}$
Answer I got = $\displaystyle \frac{7}{(2x+1)^2}$

Are my answers for the ones completed correct?
$\displaystyle \frac{3(2x+1)-2(3x+2)}{(2x+1)^2}$

When expanding the numerator remember that the -2 must be distributed to all terms inside the brackets:

$\displaystyle 3(2x+1) = 6x+3$
$\displaystyle -2(3x+2) = -6x-4$

$\displaystyle \frac{6x+3-6x-4}{(2x+1)^2} = -\frac{1}{(2x+1)^2}$
• Feb 1st 2010, 04:27 AM
jimgreenmore
Quote:

Originally Posted by e^(i*pi)
As you know the power rule for differentiation is $\displaystyle \frac{d}{dx}x^n = nx^{n-1}$. From the laws of exponents $\displaystyle \sqrt{x} = x^{1/2}$

Therefore you get $\displaystyle f'(x) = 1 + \frac{1}{2}x^{(1/2) - 1}$

$\displaystyle f'(1) = 1+\frac{1}{2\sqrt1} = \frac{3}{2}$

$\displaystyle \frac{3(2x+1)-2(3x+2)}{(2x+1)^2}$

When expanding the numerator remember that the -2 must be distributed to all terms inside the brackets:

$\displaystyle 3(2x+1) = 6x+3$
$\displaystyle -2(3x+2) = -6x-4$

$\displaystyle \frac{6x+3-6x-4}{(2x+1)^2} = -\frac{1}{(2x+1)^2}$

At the point (1,2) wasn't part of that first question. (Worried)

Did you mean $\displaystyle \frac{-1}{(2x+1)^2}$ for the second or is $\displaystyle -\frac{1}{(2x+1)^2}$ correct?
• Feb 1st 2010, 04:48 AM
e^(i*pi)
Quote:

Originally Posted by jimgreenmore
At the point (1,2) wasn't part of that first question. (Worried)

Did you mean $\displaystyle \frac{-1}{(2x+1)^2}$ for the second or is $\displaystyle -\frac{1}{(2x+1)^2}$ correct?

Was probably me misreading the question, ignore the f'(1) bit but the differentiation is still good.

For the second it doesn't matter - those expressions are equal
• Feb 1st 2010, 04:55 AM
jimgreenmore
Oh, okay. Thanks. What is $\displaystyle \frac{3}{x^2}$ differentiated?
• Feb 1st 2010, 05:43 AM
e^(i*pi)
Quote:

Originally Posted by jimgreenmore
Oh, okay. Thanks. What is $\displaystyle \frac{3}{x^2}$ differentiated?

$\displaystyle f(x) = \frac{3}{x^2}= 3x^{-2}$

$\displaystyle f'(x) = (-2)(3)x^{-2-1}$