\displaystyle \text{Question 1: } \text{If } f \text{ is defined by } f(x)=x^{2},\ \text{find } f'(2).

\displaystyle \text{Answer:}

\displaystyle \text{Given:} f(x)=x^{2}

\displaystyle \text{We know that a polynomial function is differentiable everywhere.}

\displaystyle \text{Therefore, } f(x) \text{ is differentiable at } x=2.

\displaystyle f'(2)=\lim_{h\to 0}\frac{f(2+h)-f(2)}{h}

\displaystyle \Rightarrow f'(2)=\lim_{h\to 0}\frac{(2+h)^2-2^2}{h}

\displaystyle \Rightarrow f'(2)=\lim_{h\to 0}\frac{4+4h+h^2-4}{h}

\displaystyle \Rightarrow f'(2)=\lim_{h\to 0}\frac{h(4+h)}{h}

\displaystyle \Rightarrow f'(2)=4

\displaystyle \text{Question 2: } \text{If } f \text{ is defined by } f(x)=x^{2}-4x+7,\ \text{show that } f'(5)=2f'\left(\frac{7}{2}\right).

\displaystyle \text{Answer:}

\displaystyle \text{Given:} f(x)=x^{2}-4x+7

\displaystyle \text{Clearly, since } f(x) \text{ is a polynomial function, it is differentiable for all } x.

\displaystyle \text{The derivative of } f \text{ at } x \text{ is given by:}

\displaystyle f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}

\displaystyle \Rightarrow f'(x)=\lim_{h\to 0}\frac{(x+h)^2-4(x+h)+7-(x^2-4x+7)}{h}

\displaystyle \Rightarrow f'(x)=\lim_{h\to 0}\frac{x^2+h^2+2xh-4x-4h+7-x^2+4x-7}{h}

\displaystyle \Rightarrow f'(x)=\lim_{h\to 0}\frac{h^2+2xh-4h}{h}

\displaystyle \Rightarrow f'(x)=\lim_{h\to 0}\frac{h(h+2x-4)}{h}

\displaystyle \Rightarrow f'(x)=2x-4

\displaystyle \text{Now, } f'(5)=2(5)-4=6

\displaystyle f'\!\left(\frac{7}{2}\right)=2\!\left(\frac{7}{2}\right)-4=3

\displaystyle \therefore\ f'(5)=2\times 3=2f'\!\left(\frac{7}{2}\right)

\displaystyle \text{Hence proved.}

\displaystyle \text{Question 3: } \text{Show that the derivative of the function } f \text{ given by }  f(x)=2x^{3}-9x^{2}+12x+9,\ \text{at } x=1 \text{ and } x=2 \text{ are equal.}

\displaystyle \text{Answer:}

\displaystyle \text{Given: } f(x)=2x^{3}-9x^{2}+12x+9

\displaystyle \text{Clearly, since } f(x) \text{ is a polynomial function, it is differentiable everywhere.}

\displaystyle \text{Therefore, the derivative of } f \text{ at } x \text{ is given by:}

\displaystyle f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}

\displaystyle \Rightarrow f'(x)=\lim_{h\to 0}\frac{2(x+h)^3-9(x+h)^2+12(x+h)+9-(2x^3-9x^2+12x+9)}{h}

\displaystyle \Rightarrow f'(x)=\lim_{h\to 0}\frac{2x^3+6x^2h+6xh^2+2h^3-9x^2-18xh-9h^2+12x+12h+9-2x^3+9x^2-12x-9}{h}

\displaystyle \Rightarrow f'(x)=\lim_{h\to 0}\frac{2h^3+6xh^2-9h^2+6x^2h-18xh+12h}{h}

\displaystyle \Rightarrow f'(x)=\lim_{h\to 0}\frac{h(2h^2+6xh-9h+6x^2-18x+12)}{h}

\displaystyle \Rightarrow f'(x)=6x^{2}-18x+12

\displaystyle \text{Now,}

\displaystyle f'(1)=6(1)^2-18(1)+12=6-18+12=0

\displaystyle f'(2)=6(2)^2-18(2)+12=24-36+12=0

\displaystyle \text{Hence, the derivative at } x=1 \text{ and } x=2 \text{ are equal.}

\displaystyle \text{Question 4: } \text{If for the function } \Phi(x)=\lambda x^{2}+7x-4,\ \Phi'(5)=97,\ \text{find } \lambda.

\displaystyle \text{Answer:}

\displaystyle \text{Given: } \phi(x)=\lambda x^{2}+7x-4

\displaystyle \text{Clearly, since } \phi(x) \text{ is a polynomial function, it is differentiable everywhere.}

\displaystyle \text{Therefore, the derivative of } \phi \text{ at } x \text{ is given by:}

\displaystyle \phi'(x)=\lim_{h\to 0}\frac{\phi(x+h)-\phi(x)}{h}

\displaystyle \Rightarrow \phi'(x)=\lim_{h\to 0} \frac{\lambda(x+h)^2+7(x+h)-4-(\lambda x^2+7x-4)}{h}

\displaystyle \Rightarrow \phi'(x)=\lim_{h\to 0} \frac{\lambda x^2+2\lambda xh+\lambda h^2+7x+7h-4-\lambda x^2-7x+4}{h}

\displaystyle \Rightarrow \phi'(x)=\lim_{h\to 0}\frac{\lambda h^2+2\lambda xh+7h}{h}

\displaystyle \Rightarrow \phi'(x)=\lim_{h\to 0}\frac{h(\lambda h+2\lambda x+7)}{h}

\displaystyle \Rightarrow \phi'(x)=2\lambda x+7

\displaystyle \text{Thus,}

\displaystyle \phi'(5)=2\lambda(5)+7=10\lambda+7

\displaystyle \text{Given that } \phi'(5)=97,

\displaystyle 10\lambda+7=97

\displaystyle \Rightarrow 10\lambda=90

\displaystyle \Rightarrow \lambda=9

\displaystyle \text{Question 5: } \text{If } f(x)=x^{3}+7x^{2}+8x-9,\ \text{find } f'(4).

\displaystyle \text{Answer:}

\displaystyle \text{Given: } f(x)=x^{3}+7x^{2}+8x-9

\displaystyle \text{Clearly, since } f(x) \text{ is a polynomial function, it is differentiable everywhere.}

\displaystyle \text{Therefore, the derivative of } f \text{ at } x \text{ is given by:}

\displaystyle f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}

\displaystyle \Rightarrow f'(x)=\lim_{h\to 0} \frac{(x+h)^3+7(x+h)^2+8(x+h)-9-(x^3+7x^2+8x-9)}{h}

\displaystyle \Rightarrow f'(x)=\lim_{h\to 0} \frac{x^3+3x^2h+3xh^2+h^3+7x^2+14xh+7h^2+8x+8h-9-x^3-7x^2-8x+9}{h}

\displaystyle \Rightarrow f'(x)=\lim_{h\to 0} \frac{h^3+3xh^2+7h^2+3x^2h+14xh+8h}{h}

\displaystyle \Rightarrow f'(x)=\lim_{h\to 0} \frac{h(h^2+3xh+7h+3x^2+14x+8)}{h}

\displaystyle \Rightarrow f'(x)=3x^{2}+14x+8

\displaystyle \text{Thus,}

\displaystyle f'(4)=3(4)^2+14(4)+8

\displaystyle =48+56+8

\displaystyle =112

\displaystyle \text{Question 6: } \text{Find the derivative of the function } f \text{ defined by } f(x)=mx+c \text{ at } x=0.

\displaystyle \text{Answer:}

\displaystyle \text{Given: } f(x)=mx+c

\displaystyle \text{Clearly, since } f(x) \text{ is a polynomial function, it is differentiable everywhere.}

\displaystyle \text{Therefore, the derivative of } f \text{ at } x \text{ is given by:}

\displaystyle f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}

\displaystyle \Rightarrow f'(x)=\lim_{h\to 0}\frac{m(x+h)+c-(mx+c)}{h}

\displaystyle \Rightarrow f'(x)=\lim_{h\to 0}\frac{mx+mh+c-mx-c}{h}

\displaystyle \Rightarrow f'(x)=\lim_{h\to 0}\frac{mh}{h}

\displaystyle \Rightarrow f'(x)=m

\displaystyle \text{Thus, } f'(0)=m

\displaystyle \text{Question 7: } \text{Examine the differentiability of the function } f \text{ defined by} \\ f(x)= \begin{cases} 2x+3, & -3\le x<-2 \\ x+1, & -2\le x<0 \\ x+2, & 0\le x\le 1 \end{cases}

\displaystyle \text{Answer:}

\displaystyle f(x)= \begin{cases} 2x+3, & -3\le x\le -2 \\ x+1, & -2\le x<0 \\ 1, & 0\le x\le 1 \end{cases}

\displaystyle \Rightarrow f'(x)= \begin{cases} 2, & -3<x<-2 \\ 1, & -2<x<0 \\ 0, & 0<x<1 \end{cases}

\displaystyle \text{Now,}

\displaystyle \text{LHL at } x=-2 =\lim_{x\to -2^-} f'(x) =\lim_{x\to -2^-} 2 =2

\displaystyle \text{RHL at } x=-2 =\lim_{x\to -2^+} f'(x) =\lim_{x\to -2^+} 1 =1

\displaystyle \text{Since, at } x=-2,\ \text{LHL} \ne \text{RHL}

\displaystyle \text{Hence, } f(x) \text{ is not differentiable at } x=-2.

\displaystyle \text{Again,}

\displaystyle \text{LHL at } x=0 =\lim_{x\to 0^-} f'(x) =\lim_{x\to 0^-} 1 =1

\displaystyle \text{RHL at } x=0 =\lim_{x\to 0^+} f'(x) =\lim_{x\to 0^+} 0 =0

\displaystyle \text{Since, at } x=0,\ \text{LHL} \ne \text{RHL}

\displaystyle \text{Hence, } f(x) \text{ is not differentiable at } x=0.

\displaystyle \text{Question 8: } \text{Write an example of a function which is everywhere continuous but fails} \\ \text{to be differentiable exactly at five points.}

\displaystyle \text{Answer:}

\displaystyle f(x)=|x|+|x+1|+|x+2|+|x+3|+|x+4|

\displaystyle \text{The above function is continuous everywhere but not differentiable at } \\ x=0,-1,-2,-3 \text{ and } -4.

\displaystyle \text{Question 9: } \text{Discuss the continuity and differentiability of } f(x)=|\log|x||.

\displaystyle \text{Answer:}

\displaystyle \text{We have, } f(x)=|\log|x||

\displaystyle |x|= \begin{cases} -x, & -\infty<x<0 \\ x, & 0<x<\infty \end{cases}

\displaystyle \log|x|= \begin{cases} \log(-x), & -\infty<x<0 \\ \log(x), & 0<x<\infty \end{cases}

\displaystyle |\log|x||= \begin{cases} \log(-x), & -\infty<x<-1 \\ -\log(-x), & -1<x<0 \\ -\log(x), & 0<x<1 \\ \log(x), & 1<x<\infty \end{cases}

\displaystyle \text{Differentiability at } x=-1:

\displaystyle \text{(LHD at } x=-1\text{)} =\lim_{x\to-1^-}\frac{f(x)-f(-1)}{x+1} =\lim_{h\to0}\frac{\log(1+h)-0}{-h} =-\lim_{h\to0}\frac{\log(1+h)}{h} =-1

\displaystyle \text{(RHD at } x=-1\text{)} =\lim_{x\to-1^+}\frac{f(x)-f(-1)}{x+1} =\lim_{h\to0}\frac{-\log(1-h)}{h} =1

\displaystyle \text{Since LHD}\ne\text{RHD, } f(x) \text{ is not differentiable at } x=-1.

\displaystyle \text{Differentiability at } x=1:

\displaystyle \text{(LHD at } x=1\text{)} =\lim_{x\to1^-}\frac{-\log(x)-0}{x-1} =-\lim_{h\to0}\frac{\log(1-h)}{h} =-1

\displaystyle \text{(RHD at } x=1\text{)} =\lim_{x\to1^+}\frac{\log(x)-0}{x-1} =\lim_{h\to0}\frac{\log(1+h)}{h} =1

\displaystyle \text{Since LHD}\ne\text{RHD, } f(x) \text{ is not differentiable at } x=1.

\displaystyle \text{At } x=0,\ f(x) \text{ is not defined.}

\displaystyle \text{Continuity at } x=-1:

\displaystyle \text{LHL}=\lim_{x\to-1^-}\log(-x)=\log(1)=0

\displaystyle \text{RHL}=\lim_{x\to-1^+}-\log(-x)=-\log(1)=0

\displaystyle f(-1)=0

\displaystyle \text{Hence, } f(x)=|\log|x|| \text{ is continuous at } x=-1.

\displaystyle \text{Continuity at } x=1:

\displaystyle \text{LHL}=\lim_{x\to1^-}-\log(x)=-\log(1)=0

\displaystyle \text{RHL}=\lim_{x\to1^+}\log(x)=\log(1)=0

\displaystyle f(1)=0

\displaystyle \text{Hence, } f(x)=|\log|x|| \text{ is continuous at } x=1.

\displaystyle \text{Since } f(x) \text{ is not defined at } x=0, \text{the function is not continuous at } x=0.

\displaystyle \text{Question 10: } \text{Discuss the continuity and differentiability of } f(x)=e^{|x|}.

\displaystyle \text{Answer:}

\displaystyle \text{Given: } f(x)=e^{|x|}

\displaystyle \Rightarrow f(x)= \begin{cases} e^{x}, & x\ge 0 \\ e^{-x}, & x<0 \end{cases}

\displaystyle \text{Continuity at } x=0:

\displaystyle \text{(LHL at } x=0\text{)}

\displaystyle \lim_{x\to 0^-} f(x) =\lim_{h\to 0} f(0-h) =\lim_{h\to 0} e^{-(0-h)} =\lim_{h\to 0} e^{h} =1

\displaystyle \text{(RHL at } x=0\text{)}

\displaystyle \lim_{x\to 0^+} f(x) =\lim_{h\to 0} f(0+h) =\lim_{h\to 0} e^{h} =1

\displaystyle f(0)=e^{0}=1

\displaystyle \therefore \lim_{x\to 0^-} f(x)=\lim_{x\to 0^+} f(x)=f(0)

\displaystyle \text{Hence, } f(x) \text{ is continuous at } x=0.

\displaystyle \text{Differentiability at } x=0:

\displaystyle \text{(LHD at } x=0\text{)}

\displaystyle \lim_{x\to 0^-} \frac{f(x)-f(0)}{x-0} =\lim_{h\to 0} \frac{f(0-h)-f(0)}{-h} =\lim_{h\to 0} \frac{e^{h}-1}{-h} =-\lim_{h\to 0} \frac{e^{h}-1}{h} =-1

\displaystyle \text{(RHD at } x=0\text{)}

\displaystyle \lim_{x\to 0^+} \frac{f(x)-f(0)}{x-0} =\lim_{h\to 0} \frac{f(0+h)-f(0)}{h} =\lim_{h\to 0} \frac{e^{h}-1}{h} =1

\displaystyle \text{Since LHD} \ne \text{RHD}

\displaystyle \text{Hence, } f(x)=e^{|x|} \text{ is not differentiable at } x=0.

\displaystyle \text{Question 11: } \text{Discuss the continuity and differentiability of } \\ f(x)= \begin{cases} (x-c)\cos\left(\dfrac{1}{x-c}\right), & x\ne c \\ 0, & x=c \end{cases}

\displaystyle \text{Answer:}

\displaystyle \text{Given: } f(x)= \begin{cases} (x-c)\cos\!\left(\dfrac{1}{x-c}\right), & x\ne c \\ 0, & x=c \end{cases}

\displaystyle \text{Continuity at } x=c:

\displaystyle \text{(LHL at } x=c\text{)}

\displaystyle \lim_{x\to c^-} f(x) =\lim_{h\to 0} f(c-h) =\lim_{h\to 0} (c-h-c)\cos\!\left(\dfrac{1}{c-h-c}\right)

\displaystyle =\lim_{h\to 0} (-h)\cos\!\left(\dfrac{1}{h}\right) =0

\displaystyle \text{since } \cos\!\left(\dfrac{1}{h}\right) \text{ is bounded.}

\displaystyle \text{(RHL at } x=c\text{)}

\displaystyle \lim_{x\to c^+} f(x) =\lim_{h\to 0} f(c+h) =\lim_{h\to 0} (c+h-c)\cos\!\left(\dfrac{1}{c+h-c}\right)

\displaystyle =\lim_{h\to 0} h\cos\!\left(\dfrac{1}{h}\right) =0

\displaystyle f(c)=0

\displaystyle \therefore\ \lim_{x\to c^-} f(x)=\lim_{x\to c^+} f(x)=f(c)

\displaystyle \text{Hence, } f(x) \text{ is continuous at } x=c.

\displaystyle \text{Differentiability at } x=c:

\displaystyle \text{(LHD at } x=c\text{)}

\displaystyle \lim_{x\to c^-}\frac{f(x)-f(c)}{x-c} =\lim_{h\to 0}\frac{f(c-h)-0}{-h} =\lim_{h\to 0}\frac{-h\cos\!\left(\dfrac{1}{h}\right)}{-h}

\displaystyle =\lim_{h\to 0}\cos\!\left(\dfrac{1}{h}\right)

\displaystyle \text{which oscillates between } -1 \text{ and } 1.

\displaystyle \therefore\ \text{LHD at } x=c \text{ does not exist.}

\displaystyle \text{Similarly, RHD at } x=c \text{ does not exist.}

\displaystyle \text{Hence, } f(x) \text{ is not differentiable at } x=c.

\displaystyle \text{Question 12: } \text{Is } |\sin x| \text{ differentiable? What about } \cos|x| \text{?}

\displaystyle \text{Answer:}

\displaystyle \text{Let } f(x)=|\sin x|.

\displaystyle |\sin x|= \begin{cases} -\sin x, & (2m-1)\pi < x < 2m\pi,\ m\in\mathbb{Z} \\ \sin x, & 2m\pi < x < (2m+1)\pi,\ m\in\mathbb{Z} \end{cases}

\displaystyle \text{Differentiability at } x=2m\pi:

\displaystyle \text{(LHD at } x=2m\pi\text{)} =\lim_{x\to 2m\pi^-}\frac{f(x)-f(2m\pi)}{x-2m\pi} =\lim_{h\to 0}\frac{-\sin(2m\pi-h)-0}{-h} =\lim_{h\to 0}\frac{\sin h}{-h} =-1

\displaystyle \text{(RHD at } x=2m\pi\text{)} =\lim_{x\to 2m\pi^+}\frac{f(x)-f(2m\pi)}{x-2m\pi} =\lim_{h\to 0}\frac{\sin(2m\pi+h)-0}{h} =\lim_{h\to 0}\frac{\sin h}{h} =1

\displaystyle \text{Since LHD} \ne \text{RHD, } f(x) \text{ is not differentiable at } x=2m\pi.

\displaystyle \text{Differentiability at } x=(2m+1)\pi:

\displaystyle \text{(LHD at } x=(2m+1)\pi\text{)} =\lim_{x\to (2m+1)\pi^-}\frac{f(x)-f((2m+1)\pi)}{x-(2m+1)\pi} =\lim_{h\to 0}\frac{\sin((2m+1)\pi-h)-0}{-h} =\lim_{h\to 0}\frac{\sin h}{-h} =-1

\displaystyle \text{(RHD at } x=(2m+1)\pi\text{)} =\lim_{x\to (2m+1)\pi^+}\frac{f(x)-f((2m+1)\pi)}{x-(2m+1)\pi} =\lim_{h\to 0}\frac{-\sin((2m+1)\pi+h)-0}{h} =\lim_{h\to 0}\frac{\sin h}{h} =1

\displaystyle \text{Since LHD} \ne \text{RHD, } f(x) \text{ is not differentiable at } x=(2m+1)\pi.

\displaystyle \text{From the above results, } f(x)=|\sin x| \text{ is not differentiable at } x=n\pi,\ n\in\mathbb{Z}.

\displaystyle \text{Now, consider } \cos|x|.

\displaystyle \cos|x|=\cos x \quad \text{for all } x\in\mathbb{R}.

\displaystyle \text{Since } \cos x \text{ is differentiable for all real } x,

\displaystyle \text{therefore } \cos|x| \text{ is differentiable everywhere.}

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