Author Topic: cherylim23's methods question thread :) (Read 13379 times) Tweet Share

brightsky · « **Reply #30 on:** May 02, 2010, 09:32:02 pm »

$(e^{x^2})^5 = e^{5x^2}$

Let $u = 5x^2$

$\frac{d}{du} e^u = e^u \cdot u'$

$\implies e^{5x^2} \left(\frac{d}{dx}5x^2\right) = e^{5x^2} \cdot 10x$

$\implies 10e^{5x^2} x$

Unless you mean... $\left((e^x)^2\right)^5$ ...

m@tty · « **Reply #31 on:** May 02, 2010, 09:45:22 pm »

$y=\left(e^{x^2}\right)^5=e^{5x^2}$

So use chain rule with $u=5x^2$ $\frac{du}{dx}=10x$

$y=e^u$ $\frac{dy}{du}=e^u$

$\frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dx}=e^u \cdot 10 x$

$=e^{5x^2} \cdot 10x$

Alternatively you could use the chain rule twice,

$y=u^5$ where $u=e^{x^2}$ then $u=e^v$ where $v=x^2$ .

$\frac{dy}{du}=5u^4$ ; $\frac{du}{dv}=e^v$ ; $\frac{dv}{dx}=2x$

$\frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}$

$=5u^4 \cdot e^v \cdot 2x$

Substitute back in:

$=5\left(e^{x^2}\right)^4 \cdot e^{x^2} \cdot 2x$

$=2x \cdot \left(e^{x^2}\right)^5=10x \cdot e^{5x^2}$

99.95 · « **Reply #32 on:** May 06, 2010, 01:40:38 pm »

how do i differentiate:

y=e^xsin^2x

the.watchman · « **Reply #33 on:** May 06, 2010, 01:46:21 pm »

Quote from: 99.95 on May 06, 2010, 01:40:38 pm

how do i differentiate:

y=e^xsin^2x

$\frac{dy}{dx} = e^x \cdot \frac{d}{dx}(\sin ^2 x) + \sin ^2 x \cdot \frac{d}{dx}(e^x)$

$= e^x \cdot 2\sin x \cdot \frac{d}{dx}(\sin x) + \sin ^2 x \cdot e^x$

$= e^x \cdot 2\sin x \cdot \cos x + \sin ^2 x \cdot e^x$

$= e^x \cdot \sin x (2\cos x + \sin x)$

This is a combination of the product and chain rules

|ll|lll| · « **Reply #34 on:** May 09, 2010, 01:01:36 am »

A few questions

What is the relation type of x = y^2?

8. State the range of the relation x2 + y2 > 9 when x, y is an element of R.

A.   [0, infinity)
B.   (– infinity, 0)
C.   (– infinity, 0]
D.   R
E.   (– infinity, –3) union (3, infinity)

7. The graph of a function with rule y = f(x) has one asymptote with equation x=4. The graph of the inverse function will have:
A no asymptote
B a vertical asymptote with equation x = 4
C a horizontal asymptote with equation y = 4
D a vertical asymptote with equation x = 1/4
E a horizontal asymptote with equation x = ¼

13. State the maximion with rule f (x) = 4 / ( rt (x-7) )
A R\{7
B R\{7}
C R+
D [7, infinity)
E (7, infinity)

Thanks, I'm pretty sleep now so I can't solve some of them

m@tty · « **Reply #35 on:** May 09, 2010, 01:10:23 am »

Well $x=y^2 \implies y=\pm\sqrt{x}$

So it's relation can be described as a square root in x.

Go look at the.watchman's post..

$x^2+y^2>9$ is a region involving all points outside the circle of radius 3 centered about the origin, so there are points below, above, as well as on either side of this circle. There are points of all y-values. So the range is $\mathbb{R}$ .

The inverse involves a reflection in the line y=x, swapping x and y. So an asymptote at $x=4$ becomes a horizontal asymptote at $y=4$ .

And what is the last question?? ...

I'm guessing it is the maximal domain of f...

Well, $f(x)=\frac{4}{\srqt{x-7}}$

Firstly for $\sqrt{x-7}$ to be defined $x-7\geq 0$ . But it is also the denominator, which cannot equal zero, therefore it cannot be zero. So the relevant domain is $x-7>0$ . So $x>7$ , or $x\in (7,\infty)$ .

the.watchman · « **Reply #36 on:** May 09, 2010, 08:55:04 am »

Hey m@tty, I thought the first answer might be a "one-to-many" relation, that's how I interpreted the question

Also:

Quote from: m@tty on May 09, 2010, 01:10:23 am

So $x>7$ , or $x\in (0,\infty)$ .

You must have been a bit sleepy - $x \in (7,\infty)$ ?

m@tty · « **Reply #37 on:** May 09, 2010, 02:06:30 pm »

Quote from: the.watchman on May 09, 2010, 08:55:04 am

Hey m@tty, I thought the first answer might be a "one-to-many" relation, that's how I interpreted the question

Also:
Quote from: m@tty on May 09, 2010, 01:10:23 am
So $x>7$ , or $x\in (0,\infty)$ .

You must have been a bit sleepy - $x \in (7,\infty)$ ?

Haha, yep..

|ll|lll| · « **Reply #38 on:** May 09, 2010, 03:10:58 pm »

thanks guys

And the.watchman's answer was what it was supposed to be for the type of relation qn xP

I'm a little confused with this question and the one I gave above.
The maximal domain for f(x)= rt (x+1) is best given by:
A) R+ union 0
B) R+
C) (-infinity, -1]
D) [-1, infinity)
E) [1, infinity)

The answer is D but I suppose it's wrong...?

brightsky · « **Reply #39 on:** May 09, 2010, 03:18:31 pm »

It's right.

The maximal domain is the largest range of x. Since 1 + x is under a square root sign, $1 + x \ge 0$ , so $x \ge -1$ . This translates to interval notations as [-1,infty).

brightsky · « **Reply #40 on:** May 09, 2010, 03:19:25 pm »

Quote from: cherylim23 on May 09, 2010, 03:17:03 pm

Quote from: cherylim23 on May 09, 2010, 01:01:36 am
7. The graph of a function with rule y = f(x) has one asymptote with equation x=4. The graph of the inverse function will have:

btw, know any graphs which have only one asymptote...?

$y = e^x$

|ll|lll| · « **Reply #41 on:** May 09, 2010, 03:31:18 pm »

Quote from: cherylim23 on May 09, 2010, 01:01:36 am

8. State the range of the relation x2 + y2 > 9 when x, y is an element of R.

A. [0, infinity)
B. (– infinity, 0)
C. (– infinity, 0]
D. R
E. (– infinity, –3) union (3, infinity)

Hmm, I still don't get why its not E.

brightsky · « **Reply #42 on:** May 09, 2010, 03:41:03 pm »

Technically the answer is D, isn't it? It's already stated that $x,y \in \mathbb{R}$ .

$x^2 + y^2 > 9 \implies y^2 > 9 - x^2$

Note the greater than sign. Hence y can be any number.

m@tty · « **Reply #43 on:** May 09, 2010, 04:33:34 pm »

Think about it graphically, the relation $x^2+y^2>9$ denotes all points whose distance from the origin is greater than 3. As $x^2+y^2>9 \implies \sqrt{x^2+y^2}>3$ . Now, $\sqrt{x^2+y^2}$ can be recognised as the hypotonuse of a right angled triangle. So the relation includes any point where a line from the origin is greater than 3 units in length.

As seen in the attached image, points of all values of y satisfy this requirement. Therefore the range is $\mathbb{R}$ .

|ll|lll| · « **Reply #44 on:** May 09, 2010, 04:52:46 pm »

I partially understood the front part.
But if you let x and y = 2,

2^2 + 2^2 = 8
8 < 9 and 8 is not > 9.

So the values between (2, -2) shouldn't be included?

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