1. ## Differentiate questions

Hi
The following questions i am having problem getting the correct answers. I need help where i have gone wrong.
1) $y=sin^{-1}(-2x-1)$
$\frac{dy}{dx}=\frac{-2}{\sqrt{1-(2x-1)^2}}$
$=\frac{-2}{\sqrt{1-(4x^2-1)}}$
$=\frac{-2}{\sqrt{2-4x^2}}$

Answers says its: $\frac{1}{\sqrt{x-x^2}}$

2) $y=sin^{-1}(\sqrt{x})$
$\frac{dy}{dx}=$ $\frac{1}{2\sqrt{x}} divide (\sqrt{1-x^2})$
$=\frac{1}{2\sqrt{x}(\sqrt{1-x^2})}$

Answers says its: $\frac{1}{2\sqrt{x-x^2}}$

3)Find the second derivative of $xe^{2x}.$
The first derivative is $e^{2x}(1+2x)$
The second derivative i got:
$=2e^{2x}+4xe^{2x}$
$=2e^{2x}(1+2x)$
Answers says its $4(x+1)e^{2x}$

4)If $y=2cos3x+5sin3x$, show that $\frac{d^2y}{dx^2}=-9y$
First derivative i got: $-6sin3x+15cos3x$
Now second derivative i got: $-18cos3x-45sin3x$
What did i do wrong? or what is the next step?

P.S

2. Originally Posted by Paymemoney
Hi
The following questions i am having problem getting the correct answers. I need help where i have gone wrong.
1) $y=sin^{-1}(-2x-1)$
$\frac{dy}{dx}=\frac{-2}{\sqrt{1-(2x-1)^2}}$

$\textcolor{red}{=\frac{-2}{\sqrt{1-(-2x-1)^2}}}$

Answers says its: $\frac{1}{\sqrt{x-x^2}}$

2) $y=sin^{-1}(\sqrt{x})$
$\frac{dy}{dx}=$ $\frac{1}{2\sqrt{x}} divide (\sqrt{1-x^2})$

$\textcolor{red}{\frac{1}{2\sqrt{x} \cdot \sqrt{1-x}}}$

Answers says its: $\frac{1}{2\sqrt{x-x^2}}$

3)Find the second derivative of $xe^{2x}.$
The first derivative is $e^{2x}(1+2x)$
The second derivative i got:
$=2e^{2x}+4xe^{2x}$

$\textcolor{red}{=2e^{2x}+(1+2x)2e^{2x}}$

$\textcolor{red}{=2e^{2x}+2e^{2x}+4xe^{2x}}$

$\textcolor{red}{=4e^{2x}+4xe^{2x}}$

$\textcolor{red}{4e^{2x}(1+x)}$

Answers says its $4(x+1)e^{2x}$

4)If $y=2cos3x+5sin3x$, show that $\frac{d^2y}{dx^2}=-9y$
First derivative i got: $-6sin3x+15cos3x$
Now second derivative i got: $-18cos3x-45sin3x$
What did i do wrong? or what is the next step? factor out a -9

P.S
...

3. 3)Find the second derivative of
The first derivative is
The second derivative i got:

<<<<<<<<<<<<<<

can you explain to me how did you go from $2e^{2x}+4xe^{2x}$ to $2e^{2x}+(1+2x)2e^{2x}$?

4. Originally Posted by Paymemoney
3)Find the second derivative of
The first derivative is
The second derivative i got:

<<<<<<<<<<<<<<

can you explain to me how did you go from $2e^{2x}+4xe^{2x}$ to $2e^{2x}+(1+2x)2e^{2x}$?
I did not use your second derivative.

$y' = e^{2x}(1+2x)$ is where I started.

use the product and chain rules from here as i did.

5. Originally Posted by skeeter
I did not use your second derivative.

$y' = e^{2x}(1+2x)$ is where I started.

use the product and chain rules from here as i did.
oh ok i get it now

6. So from $\frac{-2}{\sqrt{1-(-2x-1)^2}}$ this is what i done but it doesn't seem to be correct.
$
=\frac{-2}{\sqrt{1-(-2x-1)(-2x-1)}}$

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

$=\frac{-2}{\sqrt{-4x^2+4x}}$

$=\frac{-2}{\sqrt{-4x^2+4x}}$

$=\frac{-2}{-2x^2+2x}$

$=\frac{1}{x^2+x}$

7. looking back at the original problem ...

is the function $y = \sin^{-1}(2x-1)$ or $y = \sin^{-1}(-2x-1)$ as you originally posted ???

the given solution is for the first function.

8. Originally Posted by skeeter
looking back at the original problem ...

is the function $y = \sin^{-1}(2x-1)$ or $y = \sin^{-1}(-2x-1)$ as you originally posted ???

the given solution is for the first function.
its the first one $y = \sin^{-1}(2x-1)$

9. Originally Posted by Paymemoney
its the first one $y = \sin^{-1}(2x-1)$
then work it again. you need the algebra practice.

10. well i worked it out thanks for helping.

11. Originally Posted by Paymemoney
Hi
The following questions i am having problem getting the correct answers. I need help where i have gone wrong.
1) $y=sin^{-1}(-2x-1)$
$\frac{dy}{dx}=\frac{-2}{\sqrt{1-(2x-1)^2}}$
$=\frac{-2}{\sqrt{1-(4x^2-1)}}$
The square of 2x- 1 is NOT $4x^2- 1$!

$=\frac{-2}{\sqrt{2-4x^2}}$

Answers says its: $\frac{1}{\sqrt{x-x^2}}$

2) $y=sin^{-1}(\sqrt{x})$
$\frac{dy}{dx}=$ $\frac{1}{2\sqrt{x}} divide (\sqrt{1-x^2})$
$=\frac{1}{2\sqrt{x}(\sqrt{1-x^2})}$

Answers says its: $\frac{1}{2\sqrt{x-x^2}}$

3)Find the second derivative of $xe^{2x}.$
The first derivative is $e^{2x}(1+2x)$
The second derivative i got:
$=2e^{2x}+4xe^{2x}$
$=2e^{2x}(1+2x)$
Answers says its $4(x+1)e^{2x}$

4)If $y=2cos3x+5sin3x$, show that $\frac{d^2y}{dx^2}=-9y$
First derivative i got: $-6sin3x+15cos3x$
Now second derivative i got: $-18cos3x-45sin3x$
What did i do wrong? or what is the next step?

P.S