1. ## 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??

2. 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 $\displaystyle (x^2-5)$ ?

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

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

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

3. 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??
....

4. is that first one more like this

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

Does that make more sense?

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

6. Originally Posted by el123
Re:Above , Can someone tell me if im correct?
You are correct. But let me ask you a question. Is $\displaystyle (x^{1/2})^3=(x^3)^{1/2}$ ?

7. no i dont think so . Whys that seņor?

8. 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!

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

10. Originally Posted by el123
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?

11. 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.

12. 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.