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  • Provide solution RD Sharma maths class 12 chapter 10 differentiability exercise 10.1 question 6 maths textbook solution

Provide solution RD Sharma maths class 12 chapter 10 differentiability exercise 10.1 question 6 maths textbook solution

Answers (1)

Answer:

\left ( -\tan x \right )

Hint:

Use first principle formula to find the differentiation

Given:

\log \left ( \cos x \right )

Solution:

Let,

f x= \log \left ( \cos x \right )

f \left ( x+h \right )= \log \left ( \cos \left ( x+h \right ) \right )

Now, we will use formula of first principle to find the differentiation

\frac{d}{dx}\left ( f\left ( x \right ) \right )= \lim_{h\rightarrow 0}\frac{f\left ( x+h \right )-f\left ( x \right )}{h}

\frac{d}{dx}\left (\log \left ( \cos x \right ) \right )= \lim_{h\rightarrow 0}\frac{\log \left ( \cos \left ( x+h \right ) \right )-\log \left ( \cos x \right )}{h}

                   = \lim_{h\rightarrow 0}\frac{\log \left ( \frac{\cos \left ( x+h \right )}{\cos x} \right )}{h}                                      \left [ \because \log \mathit{A}\log \mathit{B= \log \frac{\mathit{A}}{\mathit{B}}} \right ]

Add and subtract 1 in the argument of log in numerator

 

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

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

Multiply and divide by = \frac{\cos \left ( x+h \right )-\cos x}{h}

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

=\lim_{h\rightarrow 0}\left ( \frac{1}{h} \right )\times \frac{\log \left ( 1+ \frac{\cos \left ( x+h \right )-\cos x}{\cos x} \right )}{ \frac{\cos \left ( x+h \right )-\cos x}{\cos x}}\times \frac{\cos \left ( x+h \right )-\cos x}{\cos x}

=\lim_{h\rightarrow 0}\left ( \frac{1}{h} \right )\times1\times \frac{\cos \left ( x+h \right )-\cos x}{\cos x}                                                    \left [ \because \lim_{h\rightarrow 0}\frac{\log \left ( 1+x \right )}{x}=1 \right ]

=\lim_{h\rightarrow 0} \frac{\cos \left ( x+h \right )-\cos x}{h\times \cos x}

                                     

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

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

Multiply the denominator and divide the denominator by 2

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

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

=\frac{-\sin \left ( \frac{2x}{2} \right )}{\cos x}

=\frac{-\sin x }{\cos x}

=-\tan x

Hence, the differentiation of \log \left ( \cos x \right )= \left ( -\tan x \right )

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