Question

Differentiate log sinx by 1+cosx

Answer 1

i) By quotient rule of log, log[sin(x)/{1 + cos(x)}] = log|sin(x)| - log|1+cos(x)| 

ii) Now differentiating, it is: 
cos(x)/sin(x) - {-sin(x)}/{1+cos(x)} = cot(x) + sin(x)/{1+cos(x)} 

However of the above, kindly note that both form of answers are not correct. 
The answer cot(x) + sin(x)/{1+cos(x)} is derivative of ln|sin(x)/{1+cos(x)}| and not the derivative of log|sin(x)/{1+cos(x)}|. 

As per general convention, log in mathematics are known as 'common logarithms' and stands for base 10; while 'ln' is known as natural logarithms and stands for base e. All differentiation laws of logarithms are to base e, that is natural logarithms. Hence, at first we must convert the given log to ln, which can be done by applying base changing law of logarithms. 

Using this, log|sin(x)/{1+cos(x)}| = ln|sin(x)/{1+cos(x)}|/ln|10| 

Hence, the correct answer is: 

Derivative of log|sin(x)/{1+cos(x)}| = {1/ln|10|}*[cot(x) + sin(x)/{1+cos(x)}]