1. ## logarithmic differentiaion

Hey guys my teacher is confusing the hell out of me with this one problem we went over in class.So here it is.

f(x) = (2x + 1)^10 * (3x +1)^20 / sinx sqrt(cosx)

2. ## logarithmic differentiaion

Hey guys my teacher is confusing the hell out of me with this one problem we went over in class.So here it is.

f(x) = (2x + 1)^10 * (3x +1)^20 / sinx sqrt(cosx)

3. Originally Posted by uniquereason81
Hey guys my teacher is confusing the hell out of me with this one problem we went over in class.So here it is.

f(x) = (2x + 1)^10 * (3x +1)^20 / sinx sqrt(cosx)

Is everything in the numerator divided by sinx(cosx)^.5?

4. yes but its sinx * square root(cosx)

5. Originally Posted by uniquereason81
Hey guys my teacher is confusing the hell out of me with this one problem we went over in class.So here it is.

f(x) = (2x + 1)^10 * (3x +1)^20 / sinx sqrt(cosx)

So it looks like this

$\displaystyle f(x) = \frac{(2x + 1)^{10}(3x +1)^{20}}{\sin{x}\sqrt{\cos{x}}}$

6. you got it

7. Originally Posted by 11rdc11
So it looks like this

$\displaystyle f(x) = \frac{(2x + 1)^{10}(3x +1)^{20}}{\sin{x}\sqrt{\cos{x}}}$
Ok so first thing we do is take ln of both sides

$\displaystyle \ln{y} = \ln{\frac{(2x + 1)^{10}(3x +1)^{20}}{\sin{x}\sqrt{\cos{x}}}}$

Next use the properties of ln to simplify the right side

$\displaystyle \ln{y} = \ln{(2x + 1)^{10}(3x+1)^{20}} -\ln{\sin{x}\sqrt{\cos{x}}}$

$\displaystyle \ln{y} = \ln{(2x+1)^{10}} + \ln{(3x+1)^{20}}-[\ln{\sin{x}} + \ln{\cos^{\frac{1}{2}}{x}}]$

$\displaystyle \ln{y} = 10\ln{(2x+1)} + 20\ln{(3x+1)}-\ln{\sin{x}} - \frac{\ln{\cos{x}}}{2}$

Can you finish up from here?

8. do u do the derivative of y on both sides and then move y to the other side

9. Originally Posted by uniquereason81
Hey guys my teacher is confusing the hell out of me with this one problem we went over in class.So here it is.

f(x) = (2x + 1)^10 * (3x +1)^20 / sinx sqrt(cosx)

Please don't double post. I'm working it for you on your other post.

10. Yep

$\displaystyle \ln{y}$ becomes $\displaystyle \frac{y'}{y}$

so $\displaystyle y' = y$(the derivative of the right side)

$\displaystyle y= \frac{(2x + 1)^{10}(3x +1)^{20}}{\sin{x}\sqrt{\cos{x}}}$

Do you need me to workout the derivative?

11. Hello, uniquereason81!

This is a messy one.
I must assume you know the basics of Logarithmic Differentiation.

$\displaystyle y \;=\;\frac{(2x + 1)^{10}(3x +1)^{20}}{\sin x \sqrt{\cos x}}$
Take logs and expand as much as possible . . .

$\displaystyle \ln(y) \;=\;\ln\left[\frac{(2x+1)^{10}(3x+1)^{20}}{\sin x \sqrt{\cos x}} \right]$

. . . .$\displaystyle = \;\ln\bigg[(2x+1)^{10}(3x+1)^{20}\bigg] - \ln\bigg[\sin x (\cos x)^{\frac{1}{2}}\bigg]$

. . . .$\displaystyle = \;\ln(2x+1)^{10} + \ln(3x+1)^{20} - \bigg[\ln(\sin x) + \ln(\cos x)^{\frac{1}{2}}\bigg]$

. . . .$\displaystyle = \;10\ln(2x+1) + 20\ln(3x+1) - \ln(\sin x) - \frac{1}{2}\ln(\cos x)$

Differentiate implicitly:

. . $\displaystyle \frac{1}{y}\cdot\frac{dy}{dx} \;=\;10\cdot\frac{2}{2x+1} + 20\cdot\frac{3}{3x+1} - \frac{\cos x}{\sin x} - \frac{1}{2}\cdot\frac{\text{-}\sin x}{\cos x}$

. . $\displaystyle \frac{1}{y}\cdot\frac{dy}{dx} \;=\;\frac{20}{2x+1} + \frac{60}{3x+1} - \cot x + \frac{1}{2}\tan x$

Multiply by $\displaystyle y\!:\quad\frac{dy}{dx} \;=\;y\left[\frac{20}{2x+1} + \frac{60}{3x+1} - \cot z + \frac{1}{2}\tan x\right]$

Therefore: . $\displaystyle \frac{dy}{dx}\;=\;\frac{(2x+1)^{10}(3x+1)^{20}}{\s in x\sqrt{\cos x}}\left[\frac{20}{2x+1} + \frac{60}{3x+1} - \cot x + \frac{1}{2}\tan x\right]$

12. yes please cause i don't think he went that far cause we ran out of time

13. http://www.mathhelpforum.com/math-he...rentiaion.html

Soroban finished it for you on your other post. If you have have any more questions feel free to ask.

14. Originally Posted by uniquereason81
Hey guys my teacher is confusing the hell out of me with this one problem we went over in class.So here it is.

f(x) = (2x + 1)^10 * (3x +1)^20 / sinx sqrt(cosx)