Derivative of ln(1+sinx)

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log(1 + sin(x))

$$\log{\left(\sin{\left(x \right)} + 1 \right)}$$

log(1 + sin(x))

Detail solution

Let $u = \sin{\left(x \right)} + 1$ .
The derivative of $\log{\left(u \right)}$ is $\frac{1}{u}$ .
Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(\sin{\left(x \right)} + 1\right)$ :
1. Differentiate $\sin{\left(x \right)} + 1$ term by term:
  1. The derivative of the constant $1$ is zero.
  2. The derivative of sine is cosine:
    
    $\frac{d}{d x} \sin{\left(x \right)} = \cos{\left(x \right)}$
  The result is: $\cos{\left(x \right)}$
The result of the chain rule is:

$\frac{\cos{\left(x \right)}}{\sin{\left(x \right)} + 1}$

The answer is:

$\frac{\cos{\left(x \right)}}{\sin{\left(x \right)} + 1}$

The graph

Plot the graph f(x)

Plot the graph f'(x)

The first derivative [src]

  cos(x)  
----------
1 + sin(x)

$$\frac{\cos{\left(x \right)}}{\sin{\left(x \right)} + 1}$$

Simplify

The second derivative [src]

 /    2              \ 
 | cos (x)           | 
-|---------- + sin(x)| 
 \1 + sin(x)         / 
-----------------------
       1 + sin(x)

$$- \frac{\sin{\left(x \right)} + \frac{\cos^{2}{\left(x \right)}}{\sin{\left(x \right)} + 1}}{\sin{\left(x \right)} + 1}$$

Simplify

The third derivative [src]

/            2                  \       
|       2*cos (x)      3*sin(x) |       
|-1 + ------------- + ----------|*cos(x)
|                 2   1 + sin(x)|       
\     (1 + sin(x))              /       
----------------------------------------
               1 + sin(x)

$$\frac{\left(-1 + \frac{3 \sin{\left(x \right)}}{\sin{\left(x \right)} + 1} + \frac{2 \cos^{2}{\left(x \right)}}{\left(\sin{\left(x \right)} + 1\right)^{2}}\right) \cos{\left(x \right)}}{\sin{\left(x \right)} + 1}$$

Simplify

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Identical expressions

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Derivative of ln(1+sinx)

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The solution