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(sin x) using first principle

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


Q1: What is the derivative of log(sinx)?

Answer: The derivative of log sinx is cotx.

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