Check my working please.

**el123** · August 14th 2009, 01:54 AM

Differentiate these.

$y = \frac{1}{\sqrt{1-5x^2}}$

$=(1-5x^2)^\frac{-1}{2}$

$=-\frac{1}{2} (1-5x^2)^\frac{-3}{2} (-10x)$

$=\frac{5x}{\sqrt{(x^2-5)}^3}$

Next

$y = x^5e^(1-3x)$

$=5x^4e^(1-3x)+(-3)(x^5)$

$=e^(1-3x)(5-3x)$

Just so you know , in that second equation , the e is supposed to be to the power of 1-3x. But latex wont let me show it for some reason. So please keep that in mind.

Can you please check those for me??

**skeeter** · August 14th 2009, 04:29 AM

Originally Posted by el123

Differentiate these.

$y = \frac{1}{\sqrt{1-5x^2}}$

$=(1-5x^2)^\frac{-1}{2}$

$=-\frac{1}{2} (1-5x^2)^\frac{-3}{2} (-10x)$

$=\frac{5x}{\sqrt{(x^2-5)}^3}$

Next

$y = x^5e^(1-3x)$

$=5x^4e^(1-3x)+(-3)(x^5)$

$=e^(1-3x)(5-3x)$

Just so you know , in that second equation , the e is supposed to be to the power of 1-3x. But latex wont let me show it for some reason. So please keep that in mind.

Can you please check those for me??

for the first derivative, how does the quantity $(1-5x^2)$ become $(x^2-5)$ ?

$y = x^5 \cdot e^{1-3x}$

$y' = x^5 \cdot -3e^{1-3x} + e^{1-3x} \cdot 5x^4$

$y' = x^4e^{1-3x}(5 - 3x)<br />$

**Amer** · August 14th 2009, 04:37 AM

Originally Posted by el123

Differentiate these.

$y = \frac{1}{\sqrt{1-5x^2}}$

$=(1-5x^2)^\frac{-1}{2}$

$=-\frac{1}{2} (1-5x^2)^\frac{-3}{2} (-10x)$

$=\frac{5x}{\sqrt{(x^2-5)}^3}$

Next

$y = x^5e^(1-3x)$

$= 5x^4e^(1-3x)+(-3)(x^5)$ to make the power in right form write e^{1-3x} not e^(1-3x)

$=e^(1-3x)(5-3x)$

Just so you know , in that second equation , the e is supposed to be to the power of 1-3x. But latex wont let me show it for some reason. So please keep that in mind.

Can you please check those for me??

....

**el123** · August 16th 2009, 02:06 PM

is that first one more like this

$\frac{5x}{\sqrt{(1-5x^2)^3}}$

Does that make more sense?

**el123** · August 16th 2009, 02:52 PM

Re:Above , Can someone tell me if im correct?

**VonNemo19** · August 16th 2009, 02:56 PM

Originally Posted by el123

Re:Above , Can someone tell me if im correct?

You are correct. But let me ask you a question. Is $(x^{1/2})^3=(x^3)^{1/2}$ ?

**el123** · August 16th 2009, 02:58 PM

no i dont think so . Whys that señor?

**VonNemo19** · August 16th 2009, 03:01 PM

Originally Posted by el123

no i dont think so . Whys that señor?

Because they are equal! That means that you had it right the first time!

**el123** · August 16th 2009, 03:07 PM

so why did someone ask how i got the first answer?"

**VonNemo19** · August 16th 2009, 03:12 PM

Originally Posted by el123

so why did someone ask how i got the first answer?"

Um, because I'm an idiot. I misread your first post. I thought the distinction lay in the laws of exponents, but I can see now that the problem was your differentiating. Sorry for the confusion.

Yeah, in the first one, you forgot to keep that $x^2$ with that $5$ throughout the process. Your second try is right though.

Sorry, dude.

But at least we learned something about exponents huh?

**el123** · August 16th 2009, 03:14 PM

its ok , im slowly compiling a mental file of how everything works , i am missing some of the math fundamentals so i find myself screwing up on basic things. Thanks for helping though.

**VonNemo19** · August 16th 2009, 03:16 PM

Originally Posted by el123

its ok , im slowly compiling a mental file of how everything works , i am missing some of the math fundamentals so i find myself screwing up on basic things. Thanks for helping though.

You are a gentleman, indeed.

Again, sorry for the trouble.

Thread: Check my working please.

LinkBack

Thread Tools

Display

Check my working please.

Similar Math Help Forum Discussions

check on working : (5 ^1/4 x 2 ^ 1/6)^2 / 5^-1/2 x 2 ^ 2/3

Please help check my working,thank you

check my working

Please check my working out

Please can someone check my working?

Search Tags