Chapter 5: Continuity and Differentiability

Ans: The given function is

$\text{y=}{{\text{x}}^{{{\text{x}}^{\text{2}}}\text{-3}}}\text{+(x-3}{{\text{)}}^{{{\text{x}}^{\text{2}}}}} $ . 

Now suppose that  $ \text{u=}{{\text{x}}^{{{\text{x}}^{\text{2}}}\text{-3}}} $  and  $ \text{v=(x-3}{{\text{)}}^{{{\text{x}}^{\text{2}}}}} $ 

Therefore,  $ \text{y=u+v} $ .

Now, differentiating both sides with respect to  $ \text{x} $  gives

$\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{du}}{\text{dx}}\text{+}\dfrac{\text{dv}}{\text{dx}} $                                                ……. (1)

Also,  $ \text{u=}{{\text{x}}^{{{\text{x}}^{\text{2}}}\text{-3}}} $ .

Take the logarithm on both sides of the equation.

$\begin{align}&\Rightarrow\text{logu=log(}{{\text{x}}^{{{\text{x}}^{\text{2}}}\text{-3}}}\text{)} \\&\Rightarrow\text{logu=(}{{\text{x}}^{\text{2}}}\text{-3)logx} \\ \end{align} $ 

Differentiating both sides with respect to  $ \text{x} $  gives

$\begin{align}&\dfrac{\text{1}}{\text{u}}\dfrac{\text{du}}{\text{dx}}\text{=logx}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{2}}}\text{-3)+(}{{\text{x}}^{\text{2}}}\text{-3)}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(logx)}\\&\Rightarrow\dfrac{\text{1}}{\text{u}}\dfrac{\text{du}}{\text{dx}}\text{=logx}\text{.2x+(}{{\text{x}}^{\text{2}}}\text{-3)}\text{.}\dfrac{\text{1}}{\text{x}} \\ \end{align} $ 

Hence,  $\dfrac{\text{du}}{\text{dx}}\text{=}{{\text{x}}^{{{\text{x}}^{\text{2}}}\text{-3}}}\text{.}\left[ \dfrac{{{\text{x}}^{\text{2}}}\text{-3}}{\text{x}}\text{+2 }\!\!\times\!\!\text{ logx} \right] $ .                            …… (2)

Again,  $ \text{v=(x-3}{{\text{)}}^{{{\text{x}}^{\text{2}}}}} $ .

Take the logarithm on both sides of the equation.

$\begin{align}&\Rightarrow\text{logv=log(x-3}{{\text{)}}^{{{\text{x}}^{\text{2}}}}}\\&\Rightarrow \text{logv=}{{\text{x}}^{\text{2}}}\text{log(x-3)} \\ \end{align} $ 

Now, differentiating both sides with respect to  $ \text{x} $  gives

$\begin{align}&\dfrac{\text{1}}{\text{u}}\text{.}\dfrac{\text{dv}}{\text{dx}}\text{=log(x-3)}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{2}}}\text{)+}{{\text{x}}^{\text{2}}}\text{.}\dfrac{\text{d}}{\text{dx}}\text{}\!\![\!\!\text{log(x-3)}\!\!]\!\!\text{}\\&\Rightarrow\dfrac{\text{1}}{\text{u}}\text{.}\dfrac{\text{dv}}{\text{dx}}\text{=log(x-3)}\text{.2x+}{{\text{x}}^{\text{2}}}\text{.}\dfrac{\text{1}}{\text{x-3}}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(x-3)}\\&\Rightarrow\dfrac{\text{dv}}{\text{dx}}\text{=v}\text{.}\left[\text{2xlog(x-3)+}\dfrac{{{\text{x}}^{\text{2}}}}{\text{x-3}}\text{.1} \right] \\ \end{align} $ 

Hence,$\dfrac{\text{dv}}{\text{dx}}\text{=(x-3}{{\text{)}}^{{{\text{x}}^{\text{2}}}}}\left[ \dfrac{{{\text{x}}^{\text{2}}}}{\text{x-3}}\text{+2xlog(x-3)} \right] $                                    …… (3)

Thus, from the equations (1), (2) and (3) we obtain

$\dfrac{\text{dy}}{\text{dx}}\text{=}{{\text{x}}^{{{\text{x}}^{\text{2}}}\text{-3}}}\left[ \dfrac{{{\text{x}}^{\text{2}}}\text{-3}}{\text{x}}\text{+2xlogx} \right]\text{+(x-3}{{\text{)}}^{{{\text{x}}^{\text{2}}}}}\left[ \dfrac{{{\text{x}}^{\text{2}}}}{\text{x-3}}\text{+2xlog(x-3)} \right] $ .

Ans: The given equations are  $ \text{y=12(1-cost),} $                                 …… (1)

and  $ \text{x=10(t-sint)} $                                                                    …… (2)

Then differentiating the equations (1) and (2) with respect to  $ \text{x} $  gives

$\begin{align}&\dfrac{\text{dx}}{\text{dt}}\text{=}\dfrac{\text{d}}{\text{dt}}\left[\text{10(t-sint)}\right]\text{=10}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dt}}\text{(t-sint)=10(1-cost)}\\&\dfrac{\text{dy}}{\text{dt}}\text{=}\dfrac{\text{d}}{\text{dt}}\left[\text{12(1-cost)}\right]\text{=12}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dt}}\text{(1-cost)=12 }\!\!\times\!\!\text{ }\left[ \text{0-(-sint)} \right]\text{=12sint} \\ \end{align} $ 

Therefore, by dividing  $ \dfrac{\text{dy}}{\text{dt}} $  by  $ \dfrac{\text{dx}}{\text{dt}} $  we have,

$\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\left(\dfrac{\text{dy}}{\text{dt}}\right)}{\left(\dfrac{\text{dx}}{\text{dt}}\right)}\text{=}\dfrac{\text{12sint}}{\text{10(1-cost)}}\text{=}\dfrac{\text{12 }\!\!\times\!\!\text{ 2sin}\dfrac{\text{t}}{\text{2}}\text{ }\!\!\times\!\!\text{ cos}\dfrac{\text{t}}{\text{2}}}{\text{10}\!\!\times\!\!\text{2si}{{\text{n}}^{\text{2}}}\dfrac{\text{t}}{\text{2}}} $

Hence,  $\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{6}}{\text{5}}\text{cot}\dfrac{\text{t}}{\text{2}} $ .

Ans: The given equation is  $\text{y=si}{{\text{n}}^{\text{-1}}}\text{x+si}{{\text{n}}^{\text{-1}}}\sqrt{\text{1-}{{\text{x}}^{\text{2}}}} $ .

Differentiating both sides of the equation with respect to  $ \text{x} $  gives

$\begin{align}&\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\left[\text{si}{{\text{n}}^{\text{-1}}}\text{x+si}{{\text{n}}^{\text{-1}}}\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}\right]\\&\Rightarrow\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(si}{{\text{n}}^{\text{-1}}}\text{x)+}\dfrac{\text{d}}{\text{dx}}\left(\text{si}{{\text{n}}^{\text{-1}}}\sqrt{\text{1-}{{\text{x}}^{\text{2}}}} \right) \\ \end{align} $ 

$\begin{align}&\Rightarrow\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{1}}{\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}}\text{+}\dfrac{\text{1}}{\sqrt{\text{1-}{{\left( \sqrt{\text{1-}{{\text{x}}^{\text{2}}}} \right)}^{2}}}}\text{}\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\left( \sqrt{\text{1-}{{\text{x}}^{\text{2}}}} \right) \\  &\Rightarrow\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{1}}{\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}}\text{-}\dfrac{1}{\sqrt{1-1+{{\text{x}}^{2}}}}\dfrac{\text{1}}{\text{2 }\!\!\times\!\!\text{ }\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}}\times 2\text{x}\\&\Rightarrow\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{1}}{\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}}\text{-}\dfrac{\text{1}}{\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}} \\ \end{align} $ 

Hence,  $ \dfrac{\text{dy}}{\text{dx}}\text{=0} $ .

Ans: The given equation is

 $ \begin{align}& \text{x}\sqrt{\text{1+y}}\text{+y}\sqrt{\text{1+x}}\text{=0} \\  & \Rightarrow \text{x}\sqrt{\text{1+y}}\text{=y}\sqrt{\text{1+x}} \\ \end{align} $ 

Now, squaring both sides of the equation gives

 $ \begin{align}& {{\text{x}}^{\text{2}}}\text{(1+y)=}{{\text{y}}^{\text{2}}}\text{(1+x)} \\  & \Rightarrow {{\text{x}}^{\text{2}}}\text{+}{{\text{x}}^{\text{2}}}\text{y=}{{\text{y}}^{\text{2}}}\text{+x}{{\text{y}}^{\text{2}}} \\  & \Rightarrow {{\text{x}}^{\text{2}}}\text{-}{{\text{y}}^{\text{2}}}\text{=x}{{\text{y}}^{\text{2}}}\text{-}{{\text{x}}^{\text{2}}}\text{y} \\  & \Rightarrow {{\text{x}}^{\text{2}}}\text{-}{{\text{y}}^{\text{2}}}\text{=xy(y-x)} \\  & \Rightarrow \text{(x+y)(x-y)=xy(y-x)} \\  & \therefore \text{x+y=-xy} \\  & \Rightarrow \text{(1+x)y=-x} \\  & \Rightarrow \text{y=}\dfrac{\text{-x}}{\text{(1+x)}} \\ \end{align} $ 

Now, differentiating both sides of the equation with respect to  $ \text{x} $  gives

$\dfrac{\text{dy}}{\text{dx}}\text{=-}\dfrac{\text{(1+x)}\dfrac{\text{d}}{\text{dx}}\text{(x)-x}\dfrac{\text{d}}{\text{dx}}\text{(1+x)}}{{{\text{(1+x)}}^{\text{2}}}}\text{=-}\dfrac{\text{(1+x)-x}}{{{\text{(1+x)}}^{\text{2}}}} $ 

Hence,  $ \dfrac{\text{dy}}{\text{dx}}\text{=-}\dfrac{\text{1}}{{{\text{(1+x)}}^{\text{2}}}} $.

Ans: The given equation is  ${{\text{(x-a)}}^{\text{2}}}\text{+(y-b}{{\text{)}}^{\text{2}}}\text{=}{{\text{c}}^{\text{2}}} $  .

Differentiating both sides of the equation with respect to  $ \text{x} $  gives  $ \begin{align}& \dfrac{\text{d}}{\text{dx}}\text{= }\!\![\!\!\text{ (x-a}{{\text{)}}^{\text{2}}}\text{ }\!\!]\!\!\text{ +}\dfrac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ (y-b}{{\text{)}}^{\text{2}}}\text{ }\!\!]\!\!\text{ =}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{c}}^{\text{2}}}\text{)} \\  & \Rightarrow \text{2(x-a)}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(x-a)+2(y-b)}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(y-b)=0} \\ & \Rightarrow \text{2(x-a)}\text{.1+2(y-b)}\text{.}\dfrac{\text{dy}}{\text{dx}}\text{=0} \\ \end{align} $ 

Hence,  $ \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{-(x-a)}}{\text{y-b}} $                                                  ..…... (1)

Again, differentiating both sides of the equation with respect to  $ \text{x} $  gives

 $ \begin{align}& \dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\dfrac{\text{d}}{\text{dx}}\left[ \dfrac{\text{-(x-a)}}{\text{y-b}} \right] \\ & \text{=-}\dfrac{\left[ \text{(y-b)}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(x-a)-(x-a)}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(y-b)} \right]}{{{\text{(y-b)}}^{\text{2}}}} \\ & \text{=-}\left[ \dfrac{\text{(y-b)-(x-a)}\text{.}\dfrac{\text{dy}}{\text{dx}}}{{{\text{(y-b)}}^{\text{2}}}} \right] \\  & \text{=-}\left[ \dfrac{\text{(y-b)-(x-a)}\text{.}\left\{ \dfrac{\text{-(x-a)}}{\text{y-b}} \right\}}{{{\text{(y-b)}}^{\text{2}}}} \right] \\  & \text{=-}\left[ \dfrac{{{\text{(y-b)}}^{\text{2}}}\text{+(x+a}{{\text{)}}^{\text{2}}}}{{{\text{(y-b)}}^{\text{2}}}} \right] \\ \end{align} $ 

Therefore,

 $ {{\left[ \dfrac{\text{1+}{{\left( \dfrac{\text{dy}}{\text{dx}} \right)}^{\text{2}}}}{\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}} \right]}^{\dfrac{\text{3}}{\text{2}}}}\text{=}\dfrac{{{\left[ \left( \text{1+}\dfrac{{{\text{(x-a)}}^{\text{2}}}}{{{\text{(y-b)}}^{\text{2}}}} \right) \right]}^{\dfrac{\text{3}}{\text{2}}}}}{\text{-}\left[ \dfrac{{{\text{(y-a)}}^{\text{2}}}\text{+(x-a}{{\text{)}}^{\text{2}}}}{{{\text{(y-a)}}^{\text{3}}}} \right]}\text{=}\dfrac{{{\left[ \dfrac{{{\text{(y-b)}}^{\text{2}}}\text{+(x-a}{{\text{)}}^{\text{2}}}}{{{\text{(y-b)}}^{\text{2}}}} \right]}^{\dfrac{\text{3}}{\text{2}}}}}{\text{-}\left[ \dfrac{{{\text{(y-a)}}^{\text{2}}}\text{+(x-a}{{\text{)}}^{\text{2}}}}{{{\text{(y-a)}}^{\text{3}}}} \right]}=\dfrac{{{\left[ \dfrac{{{\text{c}}^{\text{2}}}}{{{\text{(y-b)}}^{\text{2}}}} \right]}^{\dfrac{\text{3}}{\text{2}}}}}{\text{-}\dfrac{{{\text{c}}^{\text{2}}}}{{{\text{(y-b)}}^{\text{3}}}}} $ 

$\Rightarrow{{\left[\dfrac{\text{1+}{{\left(\dfrac{\text{dy}}{\text{dx}}\right)}^{\text{2}}}}{\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}}\right]}^{\dfrac{\text{3}}{\text{2}}}}\text{=}\dfrac{\dfrac{{{\text{c}}^{\text{2}}}}{{{\text{(y-b)}}^{\text{3}}}}}{\dfrac{{{\text{c}}^{\text{2}}}}{{{\text{(y-b)}}^{\text{3}}}}}=\text{-c} $ , is a constant, and is independent of  $ \text{a} $  and  $ \text{b} $ .

Ans: The given equation is  $ \text{cosy=xcos(a+y)} $ .

Then, differentiating both sides of the equation with respect to  $ \text{x} $  gives

$\begin{align}&\dfrac{\text{d}}{\text{dx}}\text{}\!\![\!\!\text{cosy}\!\!]\!\!\text{=}\dfrac{\text{d}}{\text{dx}}\text{}\!\![\!\!\text{xcos(a+y)}\!\!]\!\!\text{}\\&\Rightarrow\text{-siny}\dfrac{\text{dy}}{\text{dx}}\text{=cos(a+y)}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(x)+x}\text{.}\dfrac{\text{d}}{\text{dx}}\text{}\!\![\!\!\text{cos(a+y)}\!\!]\!\!\text{}\\&\Rightarrow\text{-siny}\dfrac{\text{dy}}{\text{dx}}\text{=cos(a+y)+x}\text{.}\!\![\!\!\text{-sin(a+y)}\!\!]\!\!\text{}\dfrac{\text{dy}}{\text{dx}} \\ \end{align} $ 

 $ \Rightarrow \text{ }\!\![\!\!\text{ xsin(a+y)-siny }\!\!]\!\!\text{ }\dfrac{\text{dy}}{\text{dx}}\text{=cos(a+y)} $                                  …….. (1)

Since  $\text{cosy=xcos(a+y)}\Rightarrow\text{x=}\dfrac{\text{cosy}}{\text{cos(a+y)}} $ , so from the equation (1) gives

$\begin{align}&\left[\dfrac{\text{cosy}}{\text{cos(a+y)}}\text{.sin(a+y)-siny}\right]\dfrac{\text{dy}}{\text{dx}}\text{=cos(a+y)}\\&\Rightarrow\text{}\!\![\!\!\text{cosy}\text{.sin(a+y)-siny}\text{.cos(a+y)}\!\!]\!\!\text{}\text{.}\dfrac{\text{dy}}{\text{dx}}\text{=co}{{\text{s}}^{\text{2}}}\text{(a+y)}\\&\Rightarrow\text{sin(a+y-y)}\dfrac{\text{dy}}{\text{dx}}\text{=co}{{\text{s}}^{\text{2}}}\text{(a+y)} \\ \end{align} $ 

Hence, it has been proved that  $\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{co}{{\text{s}}^{\text{2}}}\text{(a+y)}}{\text{sina}} $ .

Ans: The given equations are

  $ \text{x=a(cost+tsint)} $                                            …… (1)

and  $ \text{y=a(sint-tcost)} $                                        …… (2)

Then, differentiating both sides of the equation (1) with respect to  $ \text{x} $  gives

$\begin{align}&\dfrac{\text{dx}}{\text{dt}}\text{=a}\left[\text{-sint+sint}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(t)+t}\text{.}\dfrac{\text{d}}{\text{dt}}\text{(sint)}\right]\\&\text{=a}\left[ \text{-sint+sint+cost} \right]\text{=atcost} \\ \end{align} $ 

Again, differentiating both sides of the equation (2) with respect to  $ \text{x} $  gives

$\begin{align}&\dfrac{\text{dy}}{\text{dt}}\text{=a}\text{.}\dfrac{\text{d}}{\text{dt}}\text{(sint-tcost)}\\&\text{a}\left[\text{cost-}\left\{\text{cost}\text{.}\dfrac{\text{d}}{\text{dt}}\text{(t)+t}\text{.}\dfrac{\text{d}}{\text{dt}}\text{(cost)}\right\}\right]\\&\text{a}\!\![\!\!\text{ cost- }\!\!\{\!\!\text{ cost-tsint }\!\!\}\!\!\text{  }\!\!]\!\!\text{=atsint}\\\end{align}$ 

Therefore,

$\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\left(\dfrac{\text{dy}}{\text{dt}}\right)}{\left( \dfrac{\text{dx}}{\text{dx}} \right)}\text{=}\dfrac{\text{atsint}}{\text{atcost}}\text{=tant}$ 

Now, differentiating both sides with respect to  $ \text{x} $  gives

$\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\dfrac{\text{d}}{\text{dx}}\left(\dfrac{\text{dy}}{\text{dx}}\right)\text{=}\dfrac{\text{d}}{\text{dx}}\text{(tant)=se}{{\text{c}}^{\text{2}}}\text{t}\text{.}\dfrac{\text{dt}}{\text{dx}}\text{=se}{{\text{c}}^{\text{2}}}\text{t}\text{.}\dfrac{\text{1}}{\text{atcost}} $       

Hence,  $\dfrac{{{\mathbf{d}}^{\mathbf{2}}}\mathbf{y}}{\mathbf{d}{{\mathbf{x}}^{\mathbf{2}}}}\text{=}\dfrac{\text{se}{{\text{c}}^{\text{3}}}\text{(t)}}{\text{at}} $ .

Ans:  Remember that, $\left|\text{x}\right|\text{=}\left\{\begin{align}&\text{x,if}\,\text{x}\ge\text{0}\\&\text{-x,if}\,\text{x0} \\ \end{align} \right\} $ 

Therefore, if  $ \text{x}\ge \text{0,} $  then  $ \text{ f(x)=}{{\left| \text{x} \right|}^{\text{3}}}\text{=}{{\text{x}}^{\text{3}}} $ .

Then,  $ \text{{f}'}\,\text{(x)=3}{{\text{x}}^{\text{2}}} $  .

Differentiating both sides with respect to  $ \text{x} $  gives

 $ \text{{{f}'}'}\,\text{(x)=6x} $ .

Now, if $\text{x0,}$then$\text{f(x)=}{{\left|\text{x}\right|}^{\text{3}}}\text{=(-}{{\text{x}}^{\text{3}}}\text{)=}{{\text{x}}^{\text{3}}} $ .

So,  $ \text{{f}'}\,\text{(x)=3}{{\text{x}}^{\text{2}}} $ .

Therefore, differentiating both sides with respect to  $ \text{x} $  gives

 $ \text{{{f}'}'}\,\text{(x)=6x} $ .

Hence, for  $ \text{f(x)=}{{\left| \text{x} \right|}^{\text{3}}}\text{,} $   $ \text{{{f}'}'}\left( \text{x} \right) $  exists for all real values of  $ \text{x} $  and is provided as

$\text{{{f}'}'(x)=}\left\{\begin{align}&\text{6x,if}\,\text{x}\ge\text{0}\\&\text{-6x,if}\,\text{x0} \\ \end{align} \right\} $

Ans:  The given sum formula is  $ \text{sin(A+B)=sinAcosB+cosAsinB} $ .

Now, differentiating both sides with respect to  $ \text{x} $  gives

$\dfrac{\text{d}}{\text{dx}}\left[\text{sin(A+B)}\right]\text{=}\dfrac{\text{d}}{\text{dx}}\text{(sinAcosB)+}\dfrac{\text{d}}{\text{dx}}\text{(cosAsinB)} $ 

$\begin{align}&\Rightarrow\text{cos(A+B)}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(A+B)=cosB}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(sinA)+sinA}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(cosB)+sinB}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(cosA)}\\&\text{+cosA}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(sinB)}\\&\Rightarrow\text{cos(A+B)}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(A+B)=cosB}\!\!\times\!\!\text{cosA}\dfrac{\text{d}}{\text{dx}}\text{+sinA(-sinB)}\dfrac{\text{dB}}{\text{dx}}\text{+sinB(-sinA)}\!\!\times\!\!\text{}\dfrac{\text{dA}}{\text{dx}}\\&\text{+cosAcosB}\dfrac{dB}{dx}\\&\Rightarrow\text{cos(A+B)}\left[\dfrac{\text{dA}}{\text{dx}}\text{+}\dfrac{\text{dB}}{\text{dx}}\right]\text{=(cosAcosB-sinAsinB)}\!\!\times\!\!\text{}\left[\dfrac{\text{dA}}{\text{dx}}\text{+}\dfrac{\text{dB}}{\text{dx}}\right]\\\end{align} $ 

Hence the required sum formula for cosines is  $ \text{cos(A+B)=cosAcosB-sinAsinB} $ .

Ans: Let take the function $\text{f}\left(\text{x}\right)=|\text{x}|+|\text{x}-1|$ 

Observe that, the function  $ \text{f} $  is continuous everywhere, but not differentiable at $ \text{x}=0 $  and  $ \text{x}=1 $ .