Question

Differentiate using first principal
$f(x) = log \: sinx$
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Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:f(x) = log \: sinx$

So,

$\rm :\longmapsto\:f(x + h) = log \: sin(x + h)$

By definition of First Principal, we have

$\rm :\longmapsto\:\boxed{ \tt{ \: f'(x) = \displaystyle\lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \: }}$

On substituting the values, we get

$\rm \: = \:\displaystyle\lim_{h \to 0} \frac{logsin(x + h) - logsinx}{h}$

We know,

$\boxed{ \tt{ \: logx - logy = log \frac{x}{y} \: }}$

So, using this identity, we get

$\rm \: = \:\displaystyle\lim_{h \to 0} \frac{log\bigg[1 + \dfrac{sin(x + h)}{sinx} - 1 \bigg]}{h}$

$\rm \: = \:\displaystyle\lim_{h \to 0} \frac{log\bigg[1 + \dfrac{sin(x + h) - sinx}{sinx} \bigg]}{h}$

$\rm \: = \:\displaystyle\lim_{h \to 0} \frac{log\bigg[1 + \dfrac{sin(x + h) - sinx}{sinx} \bigg]}{h\bigg[\dfrac{sin(x + h) - sinx}{sinx} \bigg]} \times \bigg[\dfrac{sin(x + h) - sinx}{sinx} \bigg]$

We know,

$\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{x \to 0} \frac{ log(1 + x) }{x} = 1 \: }}$

$\rm \: = \:\displaystyle\lim_{h \to 0} \frac{sin(x + h) - sinx}{h \times sinx}$

$\rm \: = \:\dfrac{1}{sinx}\displaystyle\lim_{h \to 0} \frac{sin(x + h) - sinx}{h}$

We know,

$\boxed{ \tt{ \: sinx - siny = 2sin\bigg[\dfrac{x - y}{2} \bigg]cos\bigg[\dfrac{x + y}{2} \bigg]}}$

So, using this identity, we get

$\rm \: = \:\dfrac{1}{sinx}\displaystyle\lim_{h \to 0} \frac{2 \: cos\bigg[\dfrac{x + h + x}{2} \bigg] \: sin\bigg[\dfrac{x + h - x}{2} \bigg]}{h}$

$\rm \: = \:\dfrac{2}{sinx}\displaystyle\lim_{h \to 0} \frac{\: cos\bigg[\dfrac{2x + h}{2} \bigg] \: sin\bigg[\dfrac{h}{2} \bigg]}{h}$

$\rm \: = \:\dfrac{2}{sinx}\displaystyle\lim_{h \to 0} \frac{\: cos\bigg[\dfrac{2x + h}{2} \bigg] \: sin\bigg[\dfrac{h}{2} \bigg]}{\dfrac{h}{2} \times 2}$

$\rm \: = \:\dfrac{1}{sinx}\displaystyle\lim_{h \to 0} \frac{\: cos\bigg[\dfrac{2x + h}{2} \bigg] \: sin\bigg[\dfrac{h}{2} \bigg]}{\dfrac{h}{2}}$

We know,

$\boxed{ \tt{ \: \displaystyle\lim_{h \to 0} \frac{sinh}{h} = 1 \: }}$

So, using this, we get

$\rm \: = \:\dfrac{1}{sinx}\displaystyle\lim_{h \to 0}cos\bigg[\dfrac{2x + h}{2} \bigg] \times 1$

$\rm \: = \:\dfrac{1}{sinx} \times cosx$

$\rm \: = \:\dfrac{cosx}{sinx}$

$\rm \: = \:cotx$

$\bf\implies \:\boxed{ \tt{ \: \dfrac{d}{dx}log \: sinx \: = \: cotx \: }}$

More to know :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$