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  • Provide solution RD Sharma maths class 12 chapter 10 differentiability exercise 10 point 1 question 10 maths textbook solution

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

Answers (1)

Answer:

\frac{2}{\sqrt{1-\left ( 2x+3 \right )^{2}}}

Hint:

Use first principle formula to find the differentiation

Given:

\sin ^{-1}\left ( 2x+3 \right )

Solution:

Let,

\begin{aligned} &f(x)=\sin ^{-1}(2 x+3) \\ &f(x+h)=\sin ^{-1}(2(x+h)+3)=\sin ^{-1}(2 x+2 h+3) \end{aligned}

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

\begin{aligned} &\frac{d}{d x}(f(x))=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ &\frac{d}{d x}\left(\sin ^{-1}(2 x+3)\right)=\lim _{h \rightarrow 0} \frac{\sin ^{-1}(2 x+2 h+3)-\sin ^{-1}(2 x+3)}{h} \\ &=\lim _{h \rightarrow 0} \frac{\sin ^{-1}\left[(2 x+2 h+3) \sqrt{1-(2 x+3)^{2}}-(2 x+3) \sqrt{1-(2 x+2 h+3)^{2}}\right]}{h} \end{aligned}

                                                                                                                        \left [ \because \sin ^{-1}x-\sin ^{-1}y=\sin ^{-1}\left [ x\sqrt{1-y^{2}}-y\sqrt{1-x^{2}} \right ] \right ]                          

Let,

 

\begin{aligned} &{\left[(2 x+2 h+3) \sqrt{1-(2 x+3)^{2}}-(2 x+3) \sqrt{1-(2 x+2 h+3)^{2}}\right]=z} \\ &=\lim _{h \rightarrow 0} \frac{\sin ^{-1} z}{h} \end{aligned}

Multiply and divide by z

=\lim _{h \rightarrow 0} \frac{\sin ^{-1} z}{h}\times \frac{z}{h}

=\lim _{h \rightarrow 0} \left ( 1 \right )\times \frac{z}{h}                                                 \left [ \because \lim_{h\rightarrow 0}\frac{\sin ^{-1}x}{x}=1 \right ]

=\lim _{h \rightarrow 0} \frac{z}{h}

Put the value of z now

=\lim _{h \rightarrow 0} \frac{(2 x+2 h+3) \sqrt{1-(2 x+3)^{2}}-(2 x+3) \sqrt{1-(2 x+2 h+3)^{2}}}{h}

Multiply and divide by (2 x+2 h+3) \sqrt{1-(2 x+3)^{2}}-(2 x+3) \sqrt{1-(2 x+2 h+3)^{2}}

\begin{aligned} &\quad\left[(2 x+2 h+3) \sqrt{1-(2 x+3)^{2}}-(2 x+3) \sqrt{1-(2 x+2 h+3)^{2}}\right] \\ &=\lim _{h \rightarrow 0} \frac{\times\left[(2 x+2 h+3) \sqrt{1-(2 x+3)^{2}}+(2 x+3) \sqrt{1-(2 x+2 h+3)^{2}}\right]}{h\left[(2 x+2 h+3) \sqrt{1-(2 x+3)^{2}}+(2 x+3) \sqrt{1-(2 x+2 h+3)^{2}}\right]} \\ &=\lim _{h \rightarrow 0} \frac{\left((2 x+2 h+3) \sqrt{1-(2 x+3)^{2}}\right)^{2}-\left((2 x+3) \sqrt{1-(2 x+2 h+3)^{2}}\right)^{2}}{h\left[(2 x+2 h+3) \sqrt{1-(2 x+3)^{2}}+(2 x+3) \sqrt{1-(2 x+2 h+3)^{2}}\right]} \end{aligned}

                                                                  \because \left ( a+b \right )\left ( a-b \right )=a^{2}-b^{2}

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{\left[(2 x+2 h+3)^{2}\left(1-(2 x+3)^{2}\right)\right]-\left[(2 x+3)^{2}\left(1-(2 x+2 h+3)^{2}\right)\right]}{h\left[(2 x+2 h+3) \sqrt{1-(2 x+3)^{2}}+(2 x+3) \sqrt{1-(2 x+2 h+3)^{2}}\right]} \\ &=\lim _{h \rightarrow 0} \frac{[(2 x+3)+2 h]^{2}\left[1-(2 x+3)^{2}\right]-\left[(2 x+3)^{2}\right]\left[1-((2 x+3)+2 h)^{2}\right]}{h\left[(2 x+2 h+3) \sqrt{1-(2 x+3)^{2}}+(2 x+3) \sqrt{1-(2 x+2 h+3)^{2}}\right]} \end{aligned}

=\lim _{h \rightarrow 0} \frac{\left\{\left[(2 x+3)^{2}+4 h^{2}+4 h(2 x+3)\right]\left[1-(2 x+3)^{2}\right]\right\}\left\{(2 x+3)^{2}\left(1-(2 x+3)^{2}-4 h^{2}-4 h(2 x+3)\right)\right\}}{h\left[(2 x+2 h+3) \sqrt{1-(2 x+3)^{2}}+(2 x+3) \sqrt{1-(2 x+2 h+3)^{2}}\right]}

                                                            \left ( \left ( a^{2}+b^{2}\right )=a^{2}+b^{2}+2ab \right )

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{\left[\begin{array}{l} (2 x+3)^{2}+4 h^{2}+4 h(2 x+3)-(2 x+3)^{4}-4 h^{2}(2 x+3)^{2}-4 h(2 x+3)^{3}-(2 x+3)^{2} \\ +(2 x+3)^{4}+4 h^{2}(2 x+3)^{2}+4 h(2 x+3)^{3} \end{array}\right]}{h\left[(2 x+2 h+3) \sqrt{1-(2 x+3)^{2}}+(2 x+3) \sqrt{1-(2 x+2 h+3)^{2}}\right]}\\ \end{aligned}

=\lim _{h \rightarrow 0} \frac{4 h^{2}+4 h(2 x+3)}{h\left[(2 x+2 h+3) \sqrt{1-(2 x+3)^{2}}+(2 x+3) \sqrt{1-(2 x+2 h+3)^{2}}\right]}\\ =\lim _{h \rightarrow 0} \frac{4 h(h+(2 x+3))}{h\left[(2 x+2 h+3) \sqrt{1-(2 x+3)^{2}}+(2 x+3) \sqrt{1-(2 x+2 h+3)^{2}}\right]}\\ =\lim _{h \rightarrow 0} \frac{4(h+(2 x+3))}{(2 x+2 h+3) \sqrt{1-(2 x+3)^{2}}+(2 x+3) \sqrt{1-(2 x+2 h+3)^{2}}}\\

=\frac{4(2 x+3)}{(2 x+3) \sqrt{1-(2 x+3)^{2}}+(2 x+3) \sqrt{1-(2 x+3)^{2}}}\\ =\frac{4(2 x+3)}{2 \times(2 x+3) \sqrt{1-(2 x+3)^{2}}}\\ =\frac{2}{\sqrt{1-(2 x+3)^{2}}}\\ \frac{d}{d x}\left(\sin ^{-1}(2 x+3)\right)=\frac{2}{\sqrt{1-(2 x+3)^{2}}}

Hence, the differentiation of \sin ^{-1}\left ( 2x+3 \right ) Is \frac{2}{\sqrt{1-(2 x+3)^{2}}}

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