Question

Differentiate $\log \sin x$ by first principle.

Answer 1

Hint – First principle of derivatives says that for $f(x)$, ${f^,}(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}$. So, apply the given equation to the function given in the question.

Complete step-by-step answer:
Let $f(x) = \log \sin x$
First principle of derivatives says that for $f(x)$,
${f^,}(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}$.
So, we shall apply it to the function given.
$f(x) = \log \sin x$
${f^,}(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\log (\sin (x + h)) - \log (\sin x)}}{h}$ [by first principle]
Using the addition formula, we get
$\sin (A + B) = \sin A\cos B + \cos A\sin B$.
So, the above equation will transform into-
${f^,}(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\log (\sin (x)\cos (h) + \cos (x)\sin (h)) - \log (\sin x)}}{h}$
We can also use the subtraction that says,
${\log _a}(b) - {\log _a}(c) = {\log _a}\left( {\dfrac{b}{c}} \right)$ to get, so, now the equation will be-
${f^,}(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\log \left( {\dfrac{{\sin (x)\cos (h) + \cos (x)\sin (h)}}{{\sin x}}} \right)}}{h}$
Now, dividing numerator by sin x we get-
$= \mathop {\lim }\limits_{h \to 0} \dfrac{{\log (\cos (h) + \cot (x)\sin (h))}}{h}$
Now, cos (h) = 1, as h tends to 0.
Therefore, the equation becomes-
$= \mathop {\lim }\limits_{h \to 0} \dfrac{{\log (1 + \cot (x)\sin (h))}}{h}$
Divide and multiply by $\cot (x).\sin (h)$ , we get-
$= \mathop {\lim }\limits_{h \to 0} \dfrac{{\log (1 + \cot (x)\sin (h))}}{{h(\cot (x).\sin (h))}}.\cot (x)\sin (h)$
Now using the property $\log \left( {\dfrac{{1 + x}}{x}} \right) = 1$, so this implies that-
$\dfrac{{\log (1 + \cot x.\sinh )}}{{\cot x.\sinh }} = 1$ .
Therefore, the above equation becomes-
$= \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}.\cot (x)\sin (h)$
Now we know that $\mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}\sin (h) = 1$ . So, the equation becomes now-
$= \cot (x)$
Hence, the derivative of $\log \sin x$ by first principle is cot (x).

Note- Whenever such types of question appear then always proceed using the formula ${f^,}(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}$ and be careful about evaluating limits. Just make sure that you didn’t skip any step as it is a long solution. Make the necessary assumptions when needed.