Derivative of log(sin x) by First Principle

If f(x) is a function of the real variable x, then its derivative by the first principle of the derivative is given by

$f'(x)=\lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$ $\quad \cdots (i)$

Here $’$ denotes the derivative. In this post, we will find the derivative of \log(\sin x) by the first principle of derivatives.

Derivative of log(sinx) from first principle

Let $f(x)=\log(\sin x)$.  By the above rule (i) of the first principle of derivatives, we obtain the derivative of $\log(\sin x)$ as follows:

$\dfrac{d}{dx}(\log(\sin x)) = (\log \sin x)’$$=\lim\limits_{h \to 0} \dfrac{\log(\sin(x+h))-\log(\sin x)}{h}$

Step 1: We will apply the formula of $\log m -\log n =\log (m/n)$. By doing so we get that

$=\lim\limits_{h \to 0} \dfrac{\log(\frac{\sin(x+h)}{\sin x})}{h}$

Step 2: Applying the trigonometric formula of sin(a+b)=sin a cos b+cos a sin b, the above is

$=\lim\limits_{h \to 0}$ $\dfrac{\log(\dfrac{\sin x \cos h +\cos x \sin h}{\sin x})}{h}$

$=\lim\limits_{h \to 0}$ $\dfrac{\log(\dfrac{\sin x \cos h}{\sin x} +\dfrac{\cos x \sin h}{\sin x})}{h}$

$=\lim\limits_{h \to 0}$ $\dfrac{\log(\cos h + \cot x \sin h)}{h} \quad$ as we know that $\frac{\cos x}{\sin x} =\cot x$.

Step 3: As $\cos h$ tends to $1$ when $h \to 0$, from the step 2 we obtain the derivative of $\log \sin x$ is

$=\lim\limits_{h \to 0}$ $[\dfrac{\log(1 + \cot x \sin h)}{\cot x \sin h}$ $\times \dfrac{\cot x \sin h}{h}]$

Let $t=\cot x \sin h$. Then $t \to 0$ when $h \to 0$.

$=\lim\limits_{t \to 0}$ $\dfrac{\log(1 + t)}{t}$ $\times \lim\limits_{h \to 0} \dfrac{\cot x \sin h}{h}$

$=1 \times \cot x \lim\limits_{h \to 0} \dfrac{\sin h}{h}$ as the limit of log(1+t)/t is 1 when t tends to zero.

$=\cot x \times 1$ as $\lim\limits_{h \to 0} \dfrac{\sin h}{h} =1$

$=\cot x$

Thus the derivative of log(sin x) is cot x, and this is obtained by the first principle of derivatives.

Also Read:

Derivative of root sin x from first principle

Derivative of root cos x from first principle

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