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  • Provide solution RD Sharma maths class 12 chapter 10 differentiability exercise 10.1 question 9 maths textbook solution

Provide solution RD Sharma maths class 12 chapter 10 differentiability exercise 10.1 question 9 maths textbook solution

Answers (1)

Answer:

-\cot x

Hint:

Use first principle formula to find the differentiation

Given:

\log \left ( \cos ecx \right )

Solution:

Let,

\begin{aligned} &f(x)=\log (\cos e c x) \\ &f(x+h)=\log (\cos e c(x+h)) \end{aligned}

Now, we will use the formula of first principle

\begin{aligned} &\frac{d}{d x}(f(x))=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ &\frac{d}{d x}(\log (\operatorname{cosec} x))=\lim _{h \rightarrow 0} \frac{\log (\operatorname{cosec}(x+h)-\log (\cos e c x))}{h} \end{aligned}

=\lim _{h \rightarrow 0} \frac{\log \left(\frac{\cos e c(x+h)}{\cos e c x}\right)}{h}                                                                  \left [ \because \log a-\log b=\log \left ( \frac{a}{b} \right ) \right ]

=\lim _{h \rightarrow 0} \frac{\log \left(\frac{\frac{1}{\sin (x+h)}}{\frac{1}{\sin x}}\right)}{h}                                                                       \left [ \because \cos ecx=\frac{1}{\sin x} \right ]

=\lim _{h \rightarrow 0} \frac{\log \left(\frac{\sin x}{\sin (x+h)}\right)}{h}

Add 1 and subtract 1 from argument of the numerator

 

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{\log \left(1+\frac{\sin x}{\sin (x+h)}-1\right)}{h} \\ &=\lim _{h \rightarrow 0} \frac{\log \left(1+\frac{\sin x-\sin (x+h)}{\sin (x+h)}\right)}{h} \end{aligned}

Multiply and divide by \frac{\sin x-\sin (x+h)}{\sin (x+h)}

=\lim _{h \rightarrow 0} \frac{\log \left(1+\frac{\sin x-\sin (x+h)}{\sin (x+h)}\right)}{\left[\frac{\sin x-\sin (x+h)}{\sin (x+h)}\right]} \times \frac{\left[\frac{\sin x-\sin (x+h)}{\sin (x+h)}\right]}{h}

=\lim_{h\rightarrow 0}\times \frac{\sin x-\sin (x+h)}{h\sin (x+h)}                                                   \left [ \because \lim_{h\rightarrow 0}\frac{\log \left ( 1+x \right )}{x}=1 \right ]

=\lim _{h \rightarrow 0} \frac{2 \cos \left(\frac{2 x+h}{2}\right) \times \sin \left(\frac{-h}{2}\right)}{h \cdot \sin (x+h)}                          \left [ \because \sin \mathit{A}-\sin \mathit{B}=2\cos \left ( \frac{\mathit{A+B}}{2} \right )\sin \left ( \frac{\mathit{A-B}}{2} \right ) \right ]

=\lim _{h \rightarrow 0} \frac{2 \cos \left(\frac{2 x+h}{2}\right) \times \sin \left(\frac{-h}{2}\right)}{h \cdot \sin (x+h)}

Divide and multiply the denominator by (-2)

=\lim _{h \rightarrow 0} \frac{2 \cos \left(\frac{2 x+h}{2}\right) \cdot \sin \left(\frac{-h}{2}\right)}{(-2) \times \sin (x+h)\left(\frac{-h}{2}\right)}

=\lim _{h \rightarrow 0} \frac{-\cos \left(\frac{2 x+h}{2}\right)}{\sin (x+h)} \times\left(\frac{\sin \left(\frac{-h}{2}\right)}{\left(\frac{-h}{2}\right)}\right)                                                 \left [ \lim _{h \rightarrow 0}\frac{\sin x}{x}=1 \right ]

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{-\cos \left(\frac{2 x+h}{2}\right)}{\sin (x+h)} \\ &=-\cot x \\ &\therefore \frac{d}{d x}(\log \cos e c x)=-\cot x \end{aligned}

Hence, the differentiation of -\cot x

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