Help and compare answers

**jimgreenmore** · January 28th 2010, 07:32 AM

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?

**earboth** · January 28th 2010, 11:36 AM

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.

(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.

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)

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

(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:

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

Are my answers for the ones completed correct?

No. See my remarks.

**jimgreenmore** · January 28th 2010, 12:20 PM

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.

**earboth** · January 28th 2010, 09:56 PM

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
$=\bold{\color{red}4x^3}+5x-6x^2+10+x^3+5x$
$=5x^3 + 10x - 6x^2 +10$

(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.

...

**jimgreenmore** · February 1st 2010, 02:08 AM

I'm still lost here.

**e^(i*pi)** · February 1st 2010, 02:43 AM

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 $\sqrt{x} = x^{1/2}<br />$

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

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

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:

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

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

**jimgreenmore** · February 1st 2010, 04:27 AM

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 $\sqrt{x} = x^{1/2}<br />$

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.

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

**e^(i*pi)** · February 1st 2010, 04:48 AM

Originally Posted by jimgreenmore

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

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

**jimgreenmore** · February 1st 2010, 04:55 AM

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

**e^(i*pi)** · February 1st 2010, 05:43 AM

Originally Posted by jimgreenmore

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

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

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

Thread: Help and compare answers

LinkBack

Thread Tools

Display

Help and compare answers

Similar Math Help Forum Discussions

Compare

simple problem need someone to do it so i can compare answers

Compare A ,B

compare 3 alt. w/MARR

[SOLVED] All I need are anothers answers to compare with

Search Tags