# Thread: Differentiation of logarithmic function

1. ## Differentiation of logarithmic function

Got one question from book, i try to do but my answer is wrong...
Differentiate the function with respect to A.
In(cosecA + cotA)

After differentiate cosecA+cotA i stuck.

2. Originally Posted by sanikui
Got one question from book, i try to do but my answer is wrong...
Differentiate the function with respect to A.
In(cosecA + cotA)

After differentiate cosecA+cotA i stuck.
First of all, it's $\ln(\csc{A} + \cot{A})$, which stands for Natural Logarithm. Write the inverse trig functions in terms of what you already know...

So $y = \ln(\frac{1}{\sin{A}} + \frac{\cos{A}}{\sin{A}}) = \ln(\frac{1 + \cos{A}}{\sin{A}})$.

You have a composition of functions, so you need to use the chain rule.

Let $u = \frac{1 + \cos{A}}{\sin{A}}$, so $y = \ln{u}$.

Can you go from there?

Note, to find $\frac{du}{dx}$ you'll have to use the Quotient rule.

3. Originally Posted by Prove It
First of all, it's $\ln(\csc{A} + \cot{A})$, which stands for Natural Logarithm. Write the inverse trig functions in terms of what you already know...

So $y = \ln(\frac{1}{\sin{A}} + \frac{\cos{A}}{\sin{A}}) = \ln(\frac{1 + \cos{A}}{\sin{A}})$.

Mr F says: Life might be easier going one step further ..... ${\color{red}= \ln (1 + \cos A) - \ln (\sin A)}$.

[snip]
..

4. Originally Posted by sanikui
Got one question from book, i try to do but my answer is wrong...
Differentiate the function with respect to A.
In(cosecA + cotA)

After differentiate cosecA+cotA i stuck.
I transform the cosec-function and the cot-function into:

$cosec(A)=\dfrac1{\sin(A)}$ and $\cot(A)=\dfrac1{\tan(A)}$ Then your term becomes:

$f(A) = \ln\left(\dfrac1{\sin(A)} + \dfrac1{\tan(A)}\right)$ Use chain rule to differentiate:

$f'(A) = \dfrac1{\dfrac1{\sin(A)} + \dfrac1{\tan(A)}} \cdot \left(-\dfrac1{\sin^2(A)} \cdot \cos(A) - \dfrac1{\sin^2(A)}\right)$

$f'(A)=\dfrac{\sin(A) \cdot \tan(A)}{\sin(A) + \tan(A)} \cdot \dfrac{-(\cos(A)+1)}{\sin^2(A)}$ ........ Transform $\tan(A)=\dfrac{\sin(A)}{\cos(A)}$ which will yield:

$f'(A)=\dfrac{\sin^2(A)}{\cos(A) \cdot \left(\sin(A)+\dfrac{\sin(A)}{\cos(A)}\right)} \cdot \dfrac{-(\cos(A)+1)}{\sin^2(A)}$ ...... Expand the bracket and cancel out $\sin^2(A)$

$f'(A)=\dfrac1{\sin(A) \cdot \cos(A) + \sin(A)} \cdot \dfrac{-(\cos(A)+1)}{1}$

$f'(A)=\dfrac1{\sin(A) \cdot (\cos(A) + 1)} \cdot \dfrac{-(\cos(A)+1)}{1}= -\dfrac1{\sin(A)} = - cosec(A)$

5. Originally Posted by earboth
I transform the cosec-function and the cot-function into:

$cosec(A)=\dfrac1{\sin(A)}$ and $\cot(A)=\dfrac1{\tan(A)}$ Then your term becomes:

$f(A) = \ln\left(\dfrac1{\sin(A)} + \dfrac1{\tan(A)}\right)$ Use chain rule to differentiate:

$f'(A) = \dfrac1{\dfrac1{\sin(A)} + \dfrac1{\tan(A)}} \cdot \left(-\dfrac1{\sin^2(A)} \cdot \cos(A) - \dfrac1{\sin^2(A)}\right)$

$f'(A)=\dfrac{\sin(A) \cdot \tan(A)}{\sin(A) + \tan(A)} \cdot \dfrac{-(\cos(A)+1)}{\sin^2(A)}$ ........ Transform $\tan(A)=\dfrac{\sin(A)}{\cos(A)}$ which will yield:

$f'(A)=\dfrac{\sin^2(A)}{\cos(A) \cdot \left(\sin(A)+\dfrac{\sin(A)}{\cos(A)}\right)} \cdot \dfrac{-(\cos(A)+1)}{\sin^2(A)}$ ...... Expand the bracket and cancel out $\sin^2(A)$

$f'(A)=\dfrac1{\sin(A) \cdot \cos(A) + \sin(A)} \cdot \dfrac{-(\cos(A)+1)}{1}$

$f'(A)=\dfrac1{\sin(A) \cdot (\cos(A) + 1)} \cdot \dfrac{-(\cos(A)+1)}{1}= -\dfrac1{\sin(A)} = - cosec(A)$
Heavy weather, earboth .....

Continuing on from Life might be easier going one step further:

$\frac{dy}{dA} = \frac{-\sin A}{1 + \cos A} - \frac{\cos A}{\sin A} = \frac{-\sin^2 A - \cos A - \cos^2 A}{\sin A (1 + \cos A)}$ $= \frac{-1 - \cos A}{\sin A (1 + \cos A)} = \frac{-1}{\sin A} = - \text{cosec} A$.

6. Originally Posted by mr fantastic
Heavy weather, earboth .....

Continuing on from Life might be easier going one step further:

$\frac{dy}{dA} = \frac{-\sin A}{1 + \cos A} - \frac{\cos A}{\sin A} = \frac{-\sin^2 A - \cos A - \cos^2 A}{\sin A (1 + \cos A)}$ $= \frac{-1 - \cos A}{\sin A (1 + \cos A)} = \frac{-1}{\sin A} = - \text{cosec} A$.
Correct ... but sometimes you can't see the forest because of the trees

7. Yeah, i know my mistake liaw...thank all!!