\displaystyle \textbf{Question 1. }\text{Write the value of }\lim_{x\to c}\frac{f(x)-f(c)}{x-c}.
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to c}\frac{f(x)-f(c)}{x-c}\text{ is the derivative of }f(x)\text{ at }x=c
\displaystyle \therefore \lim_{x\to c}\frac{f(x)-f(c)}{x-c}=f'(c)
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\displaystyle \textbf{Question 2. }\text{Write the value of }\lim_{x\to a}\frac{xf(a)-af(x)}{x-a}.
\displaystyle \text{Answer:}
\displaystyle \frac{xf(a)-af(x)}{x-a}=\frac{xf(a)-af(a)+af(a)-af(x)}{x-a}
\displaystyle =f(a)-a\frac{f(x)-f(a)}{x-a}
\displaystyle \therefore \lim_{x\to a}\frac{xf(a)-af(x)}{x-a}=f(a)-af'(a)
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\displaystyle \textbf{Question 3. }\text{If }x<2,\text{ then write the value of }\frac{d}{dx}\left(\sqrt{x^2-4x+4}\right).
\displaystyle \text{Answer:}
\displaystyle \sqrt{x^2-4x+4}=\sqrt{(x-2)^2}=|x-2|
\displaystyle \text{Since }x<2,\ |x-2|=2-x
\displaystyle \therefore \frac{d}{dx}\left(\sqrt{x^2-4x+4}\right)=\frac{d}{dx}(2-x)=-1
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\displaystyle \textbf{Question 4. }\text{If }\frac{\pi}{2}<x<\pi,\text{ then find }\frac{d}{dx}\left(\sqrt{\frac{1+\cos2x}{2}}\right).
\displaystyle \text{Answer:}
\displaystyle \sqrt{\frac{1+\cos2x}{2}}=\sqrt{\cos^2x}=|\cos x|
\displaystyle \text{Since }\frac{\pi}{2}<x<\pi,\ \cos x<0
\displaystyle \therefore |\cos x|=-\cos x
\displaystyle \therefore \frac{d}{dx}\left(\sqrt{\frac{1+\cos2x}{2}}\right)=\frac{d}{dx}(-\cos x)=\sin x
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\displaystyle \textbf{Question 5. }\text{Write the value of }\frac{d}{dx}(x|x|).
\displaystyle \text{Answer:}
\displaystyle x|x|=\begin{cases}x^2, & x>0 \\ -x^2, & x<0\end{cases}
\displaystyle \therefore \frac{d}{dx}(x|x|)=\begin{cases}2x, & x>0 \\ -2x, & x<0\end{cases}
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\displaystyle \textbf{Question 6. }\text{Write the value of }\frac{d}{dx}\{(x+|x|)|x|\}.
\displaystyle \text{Answer:}
\displaystyle \text{For }x<0,\ |x|=-x,\text{ so }(x+|x|)|x|=(x-x)(-x)=0
\displaystyle \text{For }x>0,\ |x|=x,\text{ so }(x+|x|)|x|=(x+x)x=2x^2
\displaystyle \therefore \frac{d}{dx}\{(x+|x|)|x|\}=\begin{cases}0, & x<0 \\ 4x, & x>0\end{cases}
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\displaystyle \textbf{Question 7. }\text{If }f(x)=\frac{x^2}{|x|},\text{ write }\frac{d}{dx}(f(x)).
\displaystyle \text{Answer:}
\displaystyle f(x)=\frac{x^2}{|x|}=|x|
\displaystyle \therefore \frac{d}{dx}(f(x))=\frac{d}{dx}(|x|)
\displaystyle =\begin{cases}1, & x>0 \\ -1, & x<0\end{cases}
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\displaystyle \textbf{Question 8. }\text{Write the value of }\frac{d}{dx}(\log |x|).
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}(\log |x|)=\frac{1}{|x|}\cdot\frac{d}{dx}(|x|)
\displaystyle =\frac{1}{|x|}\cdot\frac{x}{|x|}
\displaystyle =\frac{x}{|x|^2}
\displaystyle =\frac{x}{x^2}
\displaystyle =\frac{1}{x}
\displaystyle \therefore \frac{d}{dx}(\log |x|)=\frac{1}{x},\quad x\neq0
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\displaystyle \textbf{Question 9. }\text{If }f(x)=|x|+|x-1|,\text{ write the value of }\frac{d}{dx}(f(x)).
\displaystyle \text{Answer:}
\displaystyle f(x)=|x|+|x-1|
\displaystyle \text{For }x>1,\ f(x)=x+x-1=2x-1,\text{ so }f'(x)=2
\displaystyle \text{For }0<x<1,\ f(x)=x-(x-1)=1,\text{ so }f'(x)=0
\displaystyle \text{For }x<0,\ f(x)=-x-(x-1)=-2x+1,\text{ so }f'(x)=-2
\displaystyle \therefore \frac{d}{dx}\{f(x)\}=\begin{cases}2, & x>1 \\ 0, & 0<x<1 \\ -2, & x<0\end{cases}
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\displaystyle \textbf{Question 10. }\text{Write the value of the derivative of }f(x)=|x-1|+|x-3|\text{ at }x=2.
\displaystyle \text{Answer:}
\displaystyle \text{For }1<x<3,\ |x-1|=x-1\text{ and }|x-3|=3-x
\displaystyle \therefore f(x)=x-1+3-x=2
\displaystyle \therefore f'(x)=0
\displaystyle \therefore f'(2)=0
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\displaystyle \textbf{Question 11. }\text{If }f(1)=1,\ f'(1)=2,\text{ then write the value of }\lim_{x\to1}\frac{\sqrt{f(x)}-1}{\sqrt{x}-1}.
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to1}\frac{\sqrt{f(x)}-1}{\sqrt{x}-1}
\displaystyle =\lim_{x\to1}\frac{\frac{f(x)-1}{\sqrt{f(x)}+1}}{\frac{x-1}{\sqrt{x}+1}}
\displaystyle =\lim_{x\to1}\frac{f(x)-f(1)}{x-1}\cdot\frac{\sqrt{x}+1}{\sqrt{f(x)}+1}
\displaystyle =f'(1)\cdot\frac{2}{2}=2
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\displaystyle \textbf{Question 12. }\text{Write the derivative of }f(x)=3|2+x|\text{ at }x=-3.
\displaystyle \text{Answer:}
\displaystyle f(x)=3|x+2|
\displaystyle \text{At }x=-3,\ x+2<0,\text{ so }|x+2|=-(x+2)
\displaystyle \therefore f(x)=3[-(x+2)]=-3x-6
\displaystyle \therefore f'(-3)=-3
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\displaystyle \textbf{Question 13. }\text{If }|x|<1\text{ and }y=1+x+x^2+x^3+\cdots,\text{ then write the value of }\frac{dy}{dx}.
\displaystyle \text{Answer:}
\displaystyle y=1+x+x^2+x^3+\cdots
\displaystyle \text{This is an infinite G.P. with }a=1\text{ and }r=x
\displaystyle \therefore y=\frac{1}{1-x}
\displaystyle \therefore \frac{dy}{dx}=\frac{1}{(1-x)^2}
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\displaystyle \textbf{Question 14. }\text{If }f(x)=\log_{x^2}x^3,\text{ write the value of }f'(x).
\displaystyle \text{Answer:}
\displaystyle f(x)=\log_{x^2}x^3=\frac{\log x^3}{\log x^2}
\displaystyle =\frac{3\log x}{2\log x}=\frac{3}{2}
\displaystyle \therefore f'(x)=0
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