# Check my working please.

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• Aug 14th 2009, 01:54 AM
el123
Check my working please.
Differentiate these.

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

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

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

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

Next

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

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

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

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?? :)
• Aug 14th 2009, 04:29 AM
skeeter
Quote:

Originally Posted by el123
Differentiate these.

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

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

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

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

Next

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

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

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

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 $\displaystyle (1-5x^2)$ become $\displaystyle (x^2-5)$ ?

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

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

$\displaystyle y' = x^4e^{1-3x}(5 - 3x)$
• Aug 14th 2009, 04:37 AM
Amer
Quote:

Originally Posted by el123
Differentiate these.

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

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

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

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

Next

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

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

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

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?? :)

....
• Aug 16th 2009, 02:06 PM
el123
is that first one more like this

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

Does that make more sense?
• Aug 16th 2009, 02:52 PM
el123
Re:Above , Can someone tell me if im correct?
• Aug 16th 2009, 02:56 PM
VonNemo19
Quote:

Originally Posted by el123
Re:Above , Can someone tell me if im correct?

You are correct. But let me ask you a question. Is $\displaystyle (x^{1/2})^3=(x^3)^{1/2}$ ?
• Aug 16th 2009, 02:58 PM
el123
no i dont think so . Whys that seņor?
• Aug 16th 2009, 03:01 PM
VonNemo19
Quote:

Originally Posted by el123
no i dont think so . Whys that seņor?

Because they are equal! That means that you had it right the first time!
• Aug 16th 2009, 03:07 PM
el123
so why did someone ask how i got the first answer?"
• Aug 16th 2009, 03:12 PM
VonNemo19
Quote:

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 $\displaystyle x^2$ with that $\displaystyle 5$ throughout the process. Your second try is right though.

Sorry, dude.

But at least we learned something about exponents huh?(Worried)
• Aug 16th 2009, 03:14 PM
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.
• Aug 16th 2009, 03:16 PM
VonNemo19
Quote:

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.