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Please solve RD Sharma class 12 chapter Differentials Errors and Approximations exercise 13.1 question 1 maths textbook solution

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Answer: \Delta y=0-No change


Given: y=sinx  and  x changes from  \frac{\pi }{2}  to  \frac{22}{14}

Solution: Suppose  \chi=\frac{\pi}{2}   Therefore x+\Delta x=\frac{22}{14}

\begin{aligned} &\rightarrow \frac{\pi}{2}+\Delta x=\frac{22}{14} \\\\ &\Delta x=\frac{22}{14}-\frac{\pi}{2} \end{aligned}


→ Differentiate y with respect to x


\frac{d y}{d x}=\cos x

⇒  As we know that (\sin x)=\cos x

\therefore \frac{d y}{d x}=\cos x

⇒ when,x=\frac{\pi}{2}   we have   \frac{d y}{d x}=\cos \left(\frac{\pi}{2}\right)

\begin{aligned} &\Rightarrow\left(\frac{d y}{d x}\right) x=\frac{\pi}{2}=0 \\\\ &\Rightarrow \Delta y=\left(\frac{d y}{d x}\right) \Delta x \end{aligned}


Here     \frac{d y}{d x}=0     and   \Delta x=\frac{22}{14}-\frac{\pi}{2}


\Delta y=(0)\left(\frac{22}{14}\right)-\left(\frac{\pi}{2}\right)


\Delta y=(0)\

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