Question

Differentiate log sin x from first principles .​

Answer 1

Answer:

$★ Answer :</p><p></p><p>____________________</p><p></p><p>_____________________________</p><p></p><p>Topic : Differentiation from first principles</p><p></p><p>Question : Differentiate log sin x from first principles .</p><p></p><p>Answer : cot x</p><p></p><p>___________________</p><p></p><p>____________________________</p><p></p><p>★ Step - by - step Explanation :</p><p></p><p>Let f(x) = log sin x .Then , f(x+h) = log sin (x+h)</p><p></p><p>\begin{gathered} \qquad \qquad \qquad{ \underline{ \boxed{ \large{ \purple{ \rm{ \therefore \: \frac{d}{dx} (f(x)) = \lim_{h \to 0} \: \frac{f(x + h) - f(x)}{h}}}}}}} \: \huge{\star}\bigstar \star\\ \\ \end{gathered}∴dxd(f(x))=h→0limhf(x+h)−f(x)⋆★⋆</p><p></p><p>\begin{gathered} \qquad \qquad\large : \implies\tt \frac{d}{dx} (f(x)) = \lim_{h \to 0} \frac{log \: sin(x + h) - log \: sin \: x}{h} \\ \\ \end{gathered}:⟹dxd(f(x))=h→0limhlogsin(x+h)−logsinx</p><p></p><p>\begin{gathered}\qquad \qquad\large : \implies\tt \frac{d}{dx} (f(x)) = \lim_{h \to 0} \frac{log \bigg( \frac{sin(x + h)}{sin \: x} \bigg)}{h} \: \\ \\ \end{gathered}:⟹dxd(f(x))=h→0limhlog(sinxsin(x+h))</p><p></p><p>\begin{gathered}\qquad \qquad\large : \implies\tt \frac{d}{dx} (f(x)) = \lim_{h \to 0} \frac{log \bigg(1 + \frac{sin(x + h)}{sin \: x} - 1 \bigg) }{h} \: \\ \\ \end{gathered}:⟹dxd(f(x))=h→0limhlog(1+sinxsin(x+h)−1)</p><p></p><p>\begin{gathered}\qquad \qquad\large : \implies\tt \frac{d}{dx} (f(x)) = \lim_{h \to 0} \frac{log \bigg(1 + \frac{sin(x + h) - sin \: x}{sin \: x} \bigg) }{h} \: \\ \\ \end{gathered}:⟹dxd(f(x))=h→0limhlog(1+sinxsin(x+h)−sinx)</p><p></p><p>\begin{gathered}\qquad \qquad\large : \implies\tt \frac{d}{dx} (f(x)) = \lim_{h \to 0} \frac{log \bigg(1 + \frac{sin(x + h) - sin \: x}{sin \: x} \bigg) }{h \bigg( \frac{sin(x + h) - sin \: x}{sin \: x} \bigg) } \: \times \bigg( \frac{sin(x + h) - sin \: x}{sin \: x} \bigg) \: \\ \\ \end{gathered}:⟹dxd(f(x))=h→0limh(sinxsin(x+h)−sinx)log(1+sinxsin(x+h)−sinx)×(sinxsin(x+h)−sinx)</p><p></p><p>\begin{gathered}\qquad \qquad\large : \implies\tt \frac{d}{dx} (f(x)) = \lim_{h \to 0} \frac{log \bigg(1 + \frac{sin(x + h) - sin \: x}{sin \: x} \bigg) }{h \bigg( \frac{sin(x + h) - sin \: x}{sin \: x} \bigg) } \: \times \bigg( \frac{sin(x + h) - sin \: x}{h } \bigg) \times \frac{1}{sin \: x} \: \\ \\ \end{gathered}:⟹dxd(f(x))=h→0limh(sinxsin(x+h)−sinx)log(1+sinxsin(x+h)−sinx)×(hsin(x+h)−sinx)×sinx1</p><p></p><p>\begin{gathered}\qquad \qquad\large : \implies\tt \frac{d}{dx} (f(x)) = \lim_{h \to 0} \frac{log \bigg(1 + \frac{sin(x + h) - sin \: x}{sin \: x} \bigg) }{h \bigg( \frac{sin(x + h) - sin \: x}{sin \: x} \bigg) } \times \lim_{h \to 0} \frac{2sin \: \frac{h}{2} cos(x + \frac{h}{2} )}{ h} \times \frac{1}{sin \: x } \\ \\ \end{gathered}:⟹dxd(f(x))=h→0limh(sinxsin(x+h)−sinx)log(1+sinxsin(x+h)−sinx)×h→0limh2sin2hcos(x+2h)×sinx1</p><p></p><p>\begin{gathered}\qquad \qquad\large : \implies\tt \frac{d}{dx} (f(x)) = \lim_{h \to 0} \frac{log \bigg(1 + \frac{sin(x + h) - sin \: x}{sin \: x} \bigg) }{h \bigg( \frac{sin(x + h) - sin \: x}{sin \: x} \bigg) } \times \lim_{h \to 0} \frac{sin \: (\frac{h}{2} )cos(x + \frac{h}{2} )}{ \frac{h}{2} } \times \frac{1}{sin \: x } \\ \\ \end{gathered}:⟹dxd(f(x))=h→0limh(sinxsin(x+h)−sinx)log(1+sinxsin(x+h)−sinx)×h→0lim2hsin(2h)cos(x+2h)×sinx1</p><p></p><p>\begin{gathered}\qquad \qquad\large : \implies\tt \frac{d}{dx} (f(x)) = 1 \times cos \: x \times \frac{1}{sin \: x } \\ \\ \end{gathered}:⟹dxd(f(x))=1×cosx×sinx1</p><p></p><p>\begin{gathered}\qquad \qquad\large : \implies\tt \orange{\frac{d}{dx} (f(x)) =cot \: x} \\ \end{gathered}:⟹dxd(f(x))=cotx</p><p></p><p>$