Help and compare answers

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• January 28th 2010, 08:32 AM
jimgreenmore
Help and compare answers
Differentiate the following
1. (i) $\frac{3}{x^2}$
I struggled with these because I'm used to differentiating square roots and fractions.

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

2. Find the gradient of the curve $y = x^2 + \frac{1}{x}$ at the point (1,2)
Answer I got = 3

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

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

Are my answers for the ones completed correct?
• January 28th 2010, 12:36 PM
earboth
Quote:

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

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

Now use the chain-rule.
Quote:

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

2. Find the gradient of the curve $y = x^2 + \frac{1}{x}$ at the point (1,2)
Answer I got = 3
How did you get this result. Btw I've got 1.
Quote:

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

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

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

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

Quote:

Are my answers for the ones completed correct?
No. See my remarks.
• January 28th 2010, 01:20 PM
jimgreenmore
Here is how I did them originally:

Differentiate the following

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

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

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

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

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

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

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

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

I got:

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

Thanks for the help so far.
• January 28th 2010, 10:56 PM
earboth
Quote:

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

Differentiate the following

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

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

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

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

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

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

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

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

I got:

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

Thanks for the help so far.

...
• February 1st 2010, 03:08 AM
jimgreenmore
I'm still lost here. (Thinking)
• February 1st 2010, 03:43 AM
e^(i*pi)
Quote:

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

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

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

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

Quote:

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

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

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

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

$\frac{6x+3-6x-4}{(2x+1)^2} = -\frac{1}{(2x+1)^2}$
• February 1st 2010, 05:27 AM
jimgreenmore
Quote:

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

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

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

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

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

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

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

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

Did you mean $\frac{-1}{(2x+1)^2}$ for the second or is $-\frac{1}{(2x+1)^2}$ correct?
• February 1st 2010, 05:48 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 $\frac{-1}{(2x+1)^2}$ for the second or is $-\frac{1}{(2x+1)^2}$ 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
• February 1st 2010, 05:55 AM
jimgreenmore
Oh, okay. Thanks. What is $\frac{3}{x^2}$ differentiated?
• February 1st 2010, 06:43 AM
e^(i*pi)
Quote:

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

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

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