Chapter 5: Continuity and Differentiability

Ans: Let $f(x) = \sin \left( {{x^2} + 5} \right),u(x) = {x^2} + 5$, and $v(t) = \sin t$

Then, $(vou)(x) = v(u(x)) = v\left( {{x^2} + 5} \right)$

$ = \tan \left( {{x^2} + 5} \right) = f(x)$

Thus, $f$ is a composite of two functions.

Put $t = u(x) = {x^2} + 5$

Then, we obtain $\dfrac{{dv}}{{dt}} = \dfrac{d}{{dt}}(\sin t) = \cos t = \cos \left( {{x^2} + 5} \right)$

$\dfrac{{dt}}{{dx}} = \dfrac{d}{{dx}}\left( {{x^2} + 5} \right) = \dfrac{d}{{dx}}\left( {{x^2}} \right) + \dfrac{d}{{dx}}(5) = 2x + 0 = 2x$

Therefore, by chain rule. $\dfrac{{df}}{{dx}} = \dfrac{{dv}}{{dt}} \cdot \dfrac{{dt}}{{dx}} = \cos \left( {{x^2} + 5} \right) \times 2x$

$ = 2x\cos \left( {{x^2} + 5} \right)$

Ans: Let $f(x) = \cos (\sin x),u(x) = \sin x$, and $v(t) = \cos t$

Then, $(\operatorname{vou} )(x) = v(u(x)) = v(\sin x) = \cos (\sin x) = f(x)$

Thus, $f$ is a composite function of two functions.

Put $t = u(x) = \sin x$

$\therefore \dfrac{{dv}}{{dt}} = \dfrac{d}{{dt}}[\cos t] =  - \sin t =  - \sin (\sin x)$

$\dfrac{{dt}}{{dx}} = \dfrac{d}{{dx}}(\sin x) = \cos x$

By chain rule, $\dfrac{{df}}{{dx}},\dfrac{{dv}}{{dt}} \cdot \dfrac{{dt}}{{dx}} =  - \sin (\sin x) \cdot \cos x$

$ =  - \cos x\sin (\sin x)$

Ans: Let $f(x) = \sin (ax + b),u(x) = ax + b$, and $v(t) = \sin t$

Then, $(vou)(x) = v(u(x)) = v(ax + b) = \sin (ax + b) = f(x)$

Thus, $f$ is a composite function of two functions $u$ and $v$.

Put $t = u(x) = ax + b$

Therefore, $\dfrac{{dv}}{{dt}} = \dfrac{d}{{dt}}(\sin t) = \cos t = \cos (ax + b)$

$\dfrac{{dt}}{{dx}} = \dfrac{d}{{dx}}(ax + b) = \dfrac{d}{{dx}}(ax) + \dfrac{d}{{dx}}(b) = a + 0 = a$

Hence, by chain rule, we obtain

$\dfrac{{df}}{{dx}} = \dfrac{{dv}}{{dt}} \cdot \dfrac{{dt}}{{dx}} = \cos (ax + b) \cdot a$

$ = a\cos (ax + b)$

Ans: Let $f(x) = \sec (\tan (\sqrt x )),u(x) = \sqrt x ,v(t) = \tan t$, and $w(s) = \sec s$

Then, $(vou)(x) = w[v(u(x))] = w[v(\sqrt x )] = w(\tan \sqrt x ) = \sec (\tan \sqrt x ) = f(x)$

Thus, $f$ is a composite function of three functions, $u, v$ and $w$. Put $s = v(t) = \tan t$ and $t = u(x) = \sqrt x $

$[s = \tan t]$

Then, $\dfrac{{dw}}{{ds}} = \dfrac{d}{{ds}}(\sec s) = \sec s\tan s$

$ = \sec (\tan t) \cdot \tan (\tan t)$

$ = \sec (\tan \sqrt x ) \cdot \tan (\tan \sqrt x )\quad [t = \sqrt x ]$

$\dfrac{{ds}}{{dt}} = \dfrac{d}{{dt}}(\tan t) = {\sec ^2}t = {\sec ^2}\sqrt x $

$\dfrac{{dt}}{{dx}} = \dfrac{d}{{dx}}(\sqrt x ) = \dfrac{d}{{dx}}\left( {{x^{\frac{1}{2}}}} \right)$

$ = \dfrac{1}{2} \cdot {x^{\frac{1}{2} - 1}} = \dfrac{1}{{2\sqrt x }}$

Hence, by chain rule, we obtain $\dfrac{{dt}}{{dx}} = \dfrac{{dw}}{{ds}} \cdot \dfrac{{ds}}{{dt}} \cdot \dfrac{{dt}}{{dx}}$

$ = \sec (\tan \sqrt x ) \cdot \tan (\tan \sqrt x ) \times {\sec ^2}\sqrt x  \times \dfrac{1}{{2\sqrt x }}$

$ = \dfrac{1}{{2\sqrt x }}{\sec ^2}\sqrt x (\tan \sqrt x )\tan (\tan \sqrt x )$

$ = \dfrac{{{{\sec }^2}\sqrt x \sec (\tan \sqrt x )\tan (\tan \sqrt x )}}{{2\sqrt x }}$

Ans: The given function is $f(x) = \dfrac{{\sin (ax + b)}}{{\cos (cx + d)}} = \dfrac{{g(x)}}{{h(x)}},$where $g(x) = \sin (ax + b)$ and

$h(x) = \cos (cx + d)$

$\therefore f = \dfrac{{{g^\prime }h - g{h^\prime }}}{{{h^2}}}$

Consider $g(x) = \sin (ax + b)$

Let $u(x) = ax + b,v(t) = \sin t$

Then $(vou)(x) = v(u(x)) = v(ax + b) = \sin (ax + b) = g(x)$

$\therefore g$ is a composite function of two functions, $u$ and $v$.

Put $t = u(x) = ax + b$

$\dfrac{{dv}}{{dt}} = \dfrac{d}{{dt}}(\sin t) = \cos t = \cos (ax + b)$

$\dfrac{{dt}}{{dx}} = \dfrac{d}{{dx}}(ax + b) = \dfrac{d}{{dx}}(ax) + \dfrac{d}{{dx}}(b) = a + 0 = a$

Therefore, by chain rule, we obtain ${g^\prime } = \dfrac{{dg}}{{dx}} = \dfrac{{dv}}{{dt}} \cdot \dfrac{{dt}}{{dx}} = \cos (ax + b) \cdot a = a\cos (ax + b)$

Consider $h(x) = \cos (cx + d)$

Let $p(x) = cx + d,q(y) = \cos y$

Then, $(qop)(x) = q(p(x)) = q(cx + d) = \cos (cx + d) = h(x)$

$\therefore h$ is a composite function of two functions, $p$ and $q$.

Put $y = p(x) = cx + d$

$\dfrac{{dq}}{{dy}} = \dfrac{d}{{dy}}(\cos y) =  - \sin y =  - \sin (cx + d)$

$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}(cx + d) = \dfrac{d}{{dx}}(cx) + \dfrac{d}{{dx}}(d) = c$

Therefore, by chain rule, we obtain ${h^\prime } = \dfrac{{dh}}{{dx}} = \dfrac{{dq}}{{dy}} \cdot \dfrac{{dy}}{{dx}} =  - \sin (cx + d){\text{xc}} =  - c\sin (cx + d)$

$\therefore {f^\prime } = \dfrac{{a\cos (ax + b) \cdot \cos (cx + d) - \sin (ax + b)\{  - c\sin cx + d\} }}{{{{[\cos (cx + d)]}^2}}}$

$ = \dfrac{{a\cos (ax + b)}}{{\cos (cx + d)}} + c\sin (ax + b) \cdot \dfrac{{\sin (cx + d)}}{{\cos (cx + d)}} \times \dfrac{1}{{\cos (cx + d)}}$

$ = a\cos (ax + b)\sec (cx + d) + c\sin (ax + b)\tan (cx + d)\sec (cx + d)$

Ans: $\cos {x^3} \cdot {\sin ^2}\left( {{x^5}} \right)$

$\dfrac{d}{{dx}}\left[ {\cos {x^3} \cdot {{\sin }^2}\left( {{x^5}} \right)} \right] = {\sin ^2}\left( {{x^5}} \right){\text{x}}\dfrac{d}{{dx}}\left( {\cos {x^3}} \right) + \cos {x^3}{\text{x}}\dfrac{d}{{dx}}\left[ {{{\sin }^2}\left( {{x^5}} \right)} \right]$

$ = {\sin ^2}\left( {{x^5}} \right){\text{x}}\left( { - \sin {x^3}} \right){\text{x}}\dfrac{d}{{dx}}\left( {{x^3}} \right) + \cos {x^3} + 2\sin \left( {{x^5}} \right) \cdot \dfrac{d}{{dx}}\left[ {\sin {x^5}} \right]$

The given function is

$ = \sin {x^3}{\sin ^2}\left( {{x^5}} \right) \times 3{x^2} + 2\sin {x^5}\cos {x^3} \cdot \cos {x^5} \times \dfrac{d}{{dx}}\left( {{x^5}} \right)$

$ = 3{x^2}\sin {x^3} \cdot {\sin ^3}\left( {{x^5}} \right) + 2\sin {x^5}\cos {x^5}\cos {x^3} \cdot x5{x^4}$

$ = 10{x^4}\sin {x^5}\cos {x^5}\cos {x^3} - 3{x^2}\sin {x^3}{\sin ^2}\left( {{x^5}} \right)$

Ans: $\dfrac{d}{{dx}}\left[ {2\sqrt {\cot \left( {{x^2}} \right)} } \right]$

$ = 2 \cdot \dfrac{1}{{2\sqrt {\cot \left( {{x^2}} \right)} }} \times \dfrac{{\text{d}}}{{{\text{dx}}}}\left[ {\cot \left( {{x^2}} \right)} \right]$

$ = \sqrt {\dfrac{{\sin \left( {{x^2}} \right)}}{{\cos \left( {{x^2}} \right)}}} {\text{x}} - {\operatorname{cosec} ^2}\left( {{x^2}} \right) \times \dfrac{d}{{dx}}\left( {{x^2}} \right)$

$ = \sqrt {\dfrac{{\sin \left( {{x^2}} \right)}}{{\cos \left( {{x^2}} \right)}}}  \times \dfrac{1}{{{{\sin }^2}\left( {{x^2}} \right)}}x(2x)$

$ = \dfrac{{ - 2x}}{{\sqrt {\cos {x^2}\sqrt {\sin {x^2}\sin {x^2}} } }}$

$ = \dfrac{{ - 2\sqrt 2 x}}{{\sqrt {2\sin {x^2}\cos {x^2}} \sin {x^2}}}$

$ = \dfrac{{ - 2\sqrt 2 x}}{{\sin {x^2}\sqrt {\sin 2{x^2}} }}$

Ans: Let $f(x) = \cos (\sqrt x )$

Also, let $u(x) = \sqrt x $

And, $v(t) = \cos t$

Then, $(vou)(x) = v(u(x))$

$ = v(\sqrt x )$

$ = \cos \sqrt x $

$ = f(x)$

Clearly, $f$ is a composite function of two functions, $u$ and $v$, such that $t = u(x) = \sqrt x $

Then,

$\dfrac{{dt}}{{dx}} = \dfrac{d}{{dx}}(\sqrt x ) = \dfrac{d}{{dx}}\left( {{x^{\dfrac{1}{2}}}} \right)$

$\dfrac{1}{2}{x^{ - \dfrac{1}{2}}} = \dfrac{1}{{2\sqrt x }}$

And, $\dfrac{{dv}}{{dt}} = \dfrac{d}{{dt}}(\cos t) =  - \sin t =  - \sin \sqrt x $

By using chain rule, we obtain $\dfrac{{dt}}{{dx}} = \dfrac{{dv}}{{dt}} \cdot \dfrac{{dt}}{{dx}}$

$ =  - \sin (\sqrt x ) \cdot \dfrac{1}{{2\sqrt x }}$

$ =  - \dfrac{1}{{2\sqrt x }}\sin (\sqrt x )$

$ =  - \dfrac{{\sin (\sqrt x )}}{{2\sqrt x }}$

Ans: The given function is $f(x) = |x - 1|,x \in {\mathbf{R}}$

It is known that a function $f$ is differentiable at a point $x = c$ in its domain if both $\mathop {\lim }\limits_{k \to {0^ - }} \dfrac{{f(c + h) - f(c)}}{h}$ and $\mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{f(c + h) - f(c)}}{h}$ are finite and equal.

To check the differentiability of the given function at $x = 1$, Consider the left hand limit of $f$ at $x = 1$ $\mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{f(1 + h) - f(1)}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{f|I + h - 1||1 - 1|}}{h}$

$ = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{|h| - 0}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{ - h}}{h}\quad (h < 0 \Rightarrow |h| =  - h)$

$ =  - 1$

Consider the right hand limit of $f$ at $x = 1$

$\mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{f(1 + h) - f(1)}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{f|I + h - 1| - |1 - 1|}}{h}$

$ = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{|h| - 0}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{h}{h}\quad (h > 0 \Rightarrow |h| = h)$

$ = 1$

Since the left and right hand limits of $f$ at $x = 1$ are not equal, $f$ is not differentiable at $x = 1$

Ans: The given function $f$ is $f = (x) = [x],0 < x < 3$

It is known that a function $f$ is differentiable at a point $x = c$ in its domain if both $\mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{f(c + h) - f(c)}}{h}$ and $\mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{f(c + h) - f(c)}}{h}$ are finite and equal.

To check the differentiable of the given function at $x = 1$, consider the left hand limit of $f$ at

$x = 1$

$\mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{f(1 + h) - f(1)}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{[1 + h] - [1]}}{h}$

$ = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{0 - 1}}{h} = \mathop {\lim }\limits_{h \to {0^ - }}  = \dfrac{{ - 1}}{h} = \infty $

Consider the right hand limit of $f$ at $x = 1$ $\mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{f(1 + h) - f(1)}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{[1 + h][1]}}{h}$

$ = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{1 - 1}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} 0 = 0$

Since the left and right limits of $f$ at $x = 1$ are not equal, $f$ is not differentiable at $x = 1$

To check the differentiable of the given function at $x = 2$, consider the left hand limit of $f$ at

$x = 2$

$\mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{f(2 + h) - f(2)}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{[2 + h] - [2]}}{h}$

$ = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{1 - 2}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{ - 1}}{h} = \infty $

Consider the right hand limit of $f$ at $x = 1$ 

$\mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{f(2 + h) - f(2)}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{[2 + h] - [2]}}{h}$

$ = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{1 - 2}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} 0 = 0$

Since the left and right hand limits of $f$ at $x = 2$ are not equal, $f$ is not differentiable at $x = 2$