1. ## Differentiate questions

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

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

2)$\displaystyle y=sin^{-1}(\sqrt{x})$
$\displaystyle \frac{dy}{dx}=$$\displaystyle \frac{1}{2\sqrt{x}} divide (\sqrt{1-x^2}) \displaystyle =\frac{1}{2\sqrt{x}(\sqrt{1-x^2})} Answers says its:\displaystyle \frac{1}{2\sqrt{x-x^2}} 3)Find the second derivative of \displaystyle xe^{2x}. The first derivative is \displaystyle e^{2x}(1+2x) The second derivative i got: \displaystyle =2e^{2x}+4xe^{2x} \displaystyle =2e^{2x}(1+2x) Answers says its \displaystyle 4(x+1)e^{2x} 4)If \displaystyle y=2cos3x+5sin3x, show that\displaystyle \frac{d^2y}{dx^2}=-9y First derivative i got: \displaystyle -6sin3x+15cos3x Now second derivative i got: \displaystyle -18cos3x-45sin3x What did i do wrong? or what is the next step? P.S 2. Originally Posted by Paymemoney Hi The following questions i am having problem getting the correct answers. I need help where i have gone wrong. 1)\displaystyle y=sin^{-1}(-2x-1) \displaystyle \frac{dy}{dx}=\frac{-2}{\sqrt{1-(2x-1)^2}} \displaystyle \textcolor{red}{=\frac{-2}{\sqrt{1-(-2x-1)^2}}} Answers says its:\displaystyle \frac{1}{\sqrt{x-x^2}} 2)\displaystyle y=sin^{-1}(\sqrt{x}) \displaystyle \frac{dy}{dx}=$$\displaystyle \frac{1}{2\sqrt{x}} divide (\sqrt{1-x^2})$

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

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

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

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

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

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

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

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

4)If $\displaystyle y=2cos3x+5sin3x$, show that$\displaystyle \frac{d^2y}{dx^2}=-9y$
First derivative i got: $\displaystyle -6sin3x+15cos3x$
Now second derivative i got: $\displaystyle -18cos3x-45sin3x$
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 $\displaystyle 2e^{2x}+4xe^{2x}$ to $\displaystyle 2e^{2x}+(1+2x)2e^{2x}$?

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 $\displaystyle 2e^{2x}+4xe^{2x}$ to $\displaystyle 2e^{2x}+(1+2x)2e^{2x}$?
I did not use your second derivative.

$\displaystyle y' = e^{2x}(1+2x)$ 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.

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

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

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

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

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

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

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

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

7. looking back at the original problem ...

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

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

9. Originally Posted by Paymemoney
its the first one $\displaystyle y = \sin^{-1}(2x-1)$
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)$\displaystyle y=sin^{-1}(-2x-1)$
$\displaystyle \frac{dy}{dx}=\frac{-2}{\sqrt{1-(2x-1)^2}}$
$\displaystyle =\frac{-2}{\sqrt{1-(4x^2-1)}}$
The square of 2x- 1 is NOT $\displaystyle 4x^2- 1$!

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

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

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

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

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

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

P.S