MATHEMATICS CLASS- 12
CHAPTER-5
(CONTINUITY AND
DIFFERENTIABILITY)
Miscellaneous Exercise on Chapter 5
Q.1 Differentiate
$$ y=(3x^2-9x+5)^9 $$with respect to $x$.
Solution:
Given,
$$ y=(3x^2-9x+5)^9. $$Using the chain rule,
$$ \frac{d}{dx}(u^n)=nu^{n-1}\frac{du}{dx}, $$where
$$ u=3x^2-9x+5. $$Now,
$$ \frac{du}{dx}=6x-9. $$Therefore,
$$ \begin{aligned} \frac{dy}{dx} &= 9(3x^2-9x+5)^8(6x-9). \end{aligned} $$Taking 3 common from $(6x-9)$,
$$ \boxed{ \frac{dy}{dx} = 27(2x-3)(3x^2-9x+5)^8. } $$Q.2 Differentiate
$$ y=\sin^3x+\cos^6x $$with respect to $x$.
Solution:
Given,
$$ y=\sin^3x+\cos^6x. $$Differentiate each term separately.
For the first term,
$$ \begin{aligned} \frac{d}{dx}(\sin^3x) &= 3\sin^2x\cos x. \end{aligned} $$For the second term,
$$ \begin{aligned} \frac{d}{dx}(\cos^6x) &= 6\cos^5x(-\sin x)\\[2mm] &= -6\sin x\cos^5x. \end{aligned} $$Hence,
$$ \begin{aligned} \frac{dy}{dx} &= 3\sin^2x\cos x - 6\sin x\cos^5x. \end{aligned} $$Taking $3\sin x\cos x$ common,
$$ \boxed{ \frac{dy}{dx} = 3\sin x\cos x \left( \sin x-2\cos^4x \right). } $$Q.3 Differentiate
$$ y=(5x)^{3\cos2x} $$with respect to $x$.
Solution:
Given,
$$ y=(5x)^{3\cos2x}. $$Since both the base and the exponent are functions of $x$, we use logarithmic differentiation.
Taking logarithm on both sides,
$$ \log y = 3\cos2x\cdot\log(5x). $$Differentiating both sides,
$$ \begin{aligned} \frac1y\frac{dy}{dx} &= 3(-2\sin2x)\log(5x) + 3\cos2x\cdot\frac1x\\[2mm] &= -6\sin2x\,\log(5x) + \frac{3\cos2x}{x}. \end{aligned} $$Multiplying both sides by $y$,
$$ \begin{aligned} \frac{dy}{dx} &= (5x)^{3\cos2x} \left( -6\sin2x\,\log(5x) + \frac{3\cos2x}{x} \right). \end{aligned} $$Answer:
$$ \boxed{ \frac{dy}{dx} = (5x)^{3\cos2x} \left( \frac{3\cos2x}{x} - 6\sin2x\,\log(5x) \right). } $$Q.4 Differentiate
$$ y=\sin^{-1}(x\sqrt{x}), \qquad 0\le x\le1 $$with respect to $x$.
Solution:
Given,
$$ y=\sin^{-1}(x\sqrt{x}) =\sin^{-1}(x^{3/2}). $$Using the chain rule,
$$ \frac{d}{dx}(\sin^{-1}u) = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}, $$where
$$ u=x^{3/2}. $$Now,
$$ \frac{du}{dx} = \frac32x^{1/2} = \frac32\sqrt{x}. $$Also,
$$ u^2=x^3. $$Hence,
$$ \boxed{ \frac{dy}{dx} = \frac{3\sqrt{x}} {2\sqrt{1-x^3}}. } $$Q.5 Differentiate
$$ y= \frac{\cos^{-1}\left(\dfrac{x}{2}\right)} {\sqrt{2x+7}}, \qquad -2Solution:
Given,
$$ y= \cos^{-1}\left(\frac{x}{2}\right) (2x+7)^{-1/2}. $$Using the product rule,
$$ \frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}, $$where
$$ u=\cos^{-1}\left(\frac{x}{2}\right), \qquad v=(2x+7)^{-1/2}. $$Now,
$$ \begin{aligned} \frac{du}{dx} &= -\frac{\dfrac12} {\sqrt{1-\left(\dfrac{x}{2}\right)^2}} \\[2mm] &= -\frac1{\sqrt{4-x^2}}, \end{aligned} $$and
$$ \begin{aligned} \frac{dv}{dx} &= -\frac12(2x+7)^{-3/2}\cdot2 \\[2mm] &= -(2x+7)^{-3/2}. \end{aligned} $$Therefore,
$$ \begin{aligned} \frac{dy}{dx} &= -\frac{\cos^{-1}(x/2)} {(2x+7)^{3/2}} - \frac{1} {\sqrt{4-x^2}\sqrt{2x+7}}. \end{aligned} $$Answer:
$$ \boxed{ \frac{dy}{dx} = - \frac{\cos^{-1}(x/2)} {(2x+7)^{3/2}} - \frac{1} {\sqrt{4-x^2}\sqrt{2x+7}}. } $$Q.6 Differentiate
$$ y= \cot^{-1} \left( \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}} {\sqrt{1+\sin x}-\sqrt{1-\sin x}} \right), \qquad 0Solution:
Let
$$ A=\sqrt{1+\sin x}, \qquad B=\sqrt{1-\sin x}. $$Then
$$ y=\cot^{-1}\!\left(\frac{A+B}{A-B}\right). $$Since
$$ 0Therefore,
$$ \begin{aligned} A+B &= 2\cos\frac{x}{2}, \\[2mm] A-B &= 2\sin\frac{x}{2}. \end{aligned} $$Hence,
$$ \frac{A+B}{A-B} = \cot\frac{x}{2}. $$Thus,
$$ y = \cot^{-1} \left( \cot\frac{x}{2} \right). $$Since
$$ 0<\frac{x}{2}<\frac{\pi}{4}, $$the principal value gives
$$ y=\frac{x}{2}. $$Differentiating,
$$ \boxed{ \frac{dy}{dx} = \frac12. } $$Q.7 Differentiate
$$ y=(\log x)^{\log x}, \qquad x>1 $$with respect to $x$.
Solution:
Given,
$$ y=(\log x)^{\log x}. $$Since both the base and the exponent are functions of $x$, we use logarithmic differentiation.
Taking logarithm on both sides,
$$ \log y = \log x\cdot\log(\log x). $$Differentiating both sides,
$$ \begin{aligned} \frac1y\frac{dy}{dx} &= \frac1x\log(\log x) + \log x \left( \frac1{\log x}\cdot\frac1x \right) \\[2mm] &= \frac{\log(\log x)+1}{x}. \end{aligned} $$Multiplying both sides by $y$,
$$ \boxed{ \frac{dy}{dx} = \frac{(\log x)^{\log x}} {x} \left[ \log(\log x)+1 \right]. } $$Q.8 Differentiate
$$ y=\cos(a\cos x+b\sin x), $$where $a$ and $b$ are constants.
Solution:
Given,
$$ y=\cos(a\cos x+b\sin x). $$Using the chain rule,
$$ \frac{d}{dx}(\cos u) = -\sin u\cdot\frac{du}{dx}, $$where
$$ u=a\cos x+b\sin x. $$Now,
$$ \begin{aligned} \frac{du}{dx} &= -a\sin x+b\cos x. \end{aligned} $$Therefore,
$$ \begin{aligned} \frac{dy}{dx} &= -\sin(a\cos x+b\sin x) (-a\sin x+b\cos x) \\[2mm] &= (a\sin x-b\cos x) \sin(a\cos x+b\sin x). \end{aligned} $$Answer:
$$ \boxed{ \frac{dy}{dx} = (a\sin x-b\cos x)\, \sin(a\cos x+b\sin x). } $$Q.9 Differentiate
$$ y=(\sin x-\cos x)^{(\sin x-\cos x)}, \qquad \frac{\pi}{4}Solution:
Given,
$$ y=(\sin x-\cos x)^{(\sin x-\cos x)}. $$Since both the base and the exponent are functions of $x$, we use logarithmic differentiation.
Taking logarithm on both sides,
$$ \log y = (\sin x-\cos x)\, \log(\sin x-\cos x). $$Step 1: Differentiate both sides.
$$ \begin{aligned} \frac1y\frac{dy}{dx} &= (\cos x+\sin x)\log(\sin x-\cos x) \\[2mm] &\quad+ (\sin x-\cos x) \cdot \frac{\cos x+\sin x} {\sin x-\cos x}. \end{aligned} $$Therefore,
$$ \begin{aligned} \frac1y\frac{dy}{dx} &= (\sin x+\cos x) \left[ \log(\sin x-\cos x)+1 \right]. \end{aligned} $$Step 2: Multiply both sides by $y$.
$$ \boxed{ \frac{dy}{dx} = (\sin x-\cos x)^{(\sin x-\cos x)} (\sin x+\cos x) \left[ \log(\sin x-\cos x)+1 \right]. } $$Q.10 Differentiate
$$ y=x^x+x^a+a^x+a^a, $$where $a>0$ is a constant and $x>0$.
Solution:
Given,
$$ y=x^x+x^a+a^x+a^a. $$Differentiate each term separately.
Derivative of $x^x$:
Using logarithmic differentiation,
$$ \begin{aligned} \frac{d}{dx}(x^x) &= x^x(\log x+1). \end{aligned} $$Derivative of $x^a$:
Since $a$ is a constant,
$$ \begin{aligned} \frac{d}{dx}(x^a) &= ax^{a-1}. \end{aligned} $$Derivative of $a^x$:
$$ \begin{aligned} \frac{d}{dx}(a^x) &= a^x\log a. \end{aligned} $$Derivative of $a^a$:
Since $a^a$ is a constant,
$$ \frac{d}{dx}(a^a)=0. $$Therefore,
$$ \boxed{ \frac{dy}{dx} = x^x(\log x+1) + ax^{a-1} + a^x\log a. } $$Q.11 Differentiate
$$ y=x^{x^2-3}+(x-3)^{x^2}, \qquad x>3. $$Solution:
Given,
$$ y=x^{x^2-3}+(x-3)^{x^2}. $$Differentiate each term separately using logarithmic differentiation.
First term:
Let
$$ u=x^{x^2-3}. $$Taking logarithm on both sides,
$$ \log u=(x^2-3)\log x. $$Differentiating,
$$ \begin{aligned} \frac1u\frac{du}{dx} &= 2x\log x + (x^2-3)\frac1x\\[2mm] &= 2x\log x + x-\frac3x. \end{aligned} $$Hence,
$$ \boxed{ \frac{du}{dx} = x^{x^2-3} \left( 2x\log x + x-\frac3x \right). } $$Second term:
Let
$$ v=(x-3)^{x^2}. $$Taking logarithm,
$$ \log v=x^2\log(x-3). $$Differentiating,
$$ \begin{aligned} \frac1v\frac{dv}{dx} &= 2x\log(x-3) + x^2\cdot\frac1{x-3}. \end{aligned} $$Therefore,
$$ \boxed{ \frac{dv}{dx} = (x-3)^{x^2} \left( 2x\log(x-3) + \frac{x^2}{x-3} \right). } $$Hence,
$$ \boxed{ \begin{aligned} \frac{dy}{dx} &= x^{x^2-3} \left( 2x\log x + x-\frac3x \right) \\[2mm] &\quad+ (x-3)^{x^2} \left( 2x\log(x-3) + \frac{x^2}{x-3} \right). \end{aligned} } $$Q.12 Find
$$ \frac{dy}{dx}, $$if
$$ y=12(1-\cos t), \qquad x=10(t-\sin t), \qquad -\frac{\pi}{2}For parametric equations,
$$ \frac{dy}{dx} = \frac{\dfrac{dy}{dt}} {\dfrac{dx}{dt}}. $$Step 1: Differentiate $y$ with respect to $t$.
$$ \begin{aligned} \frac{dy}{dt} &= 12\sin t. \end{aligned} $$Step 2: Differentiate $x$ with respect to $t$.
$$ \begin{aligned} \frac{dx}{dt} &= 10(1-\cos t). \end{aligned} $$Step 3: Find $\dfrac{dy}{dx}$.
$$ \begin{aligned} \frac{dy}{dx} &= \frac{12\sin t} {10(1-\cos t)} \\[2mm] &= \frac65 \cdot \frac{\sin t}{1-\cos t}. \end{aligned} $$Using the identity
$$ \frac{\sin t}{1-\cos t} = \frac{1+\cos t}{\sin t} = \cot\frac{t}{2}, $$we obtain
$$ \boxed{ \frac{dy}{dx} = \frac65 \cot\frac{t}{2}. } $$Answer:
$$ \boxed{ \frac{dy}{dx} = \frac65 \cdot \frac{\sin t}{1-\cos t} = \frac65 \cot\frac{t}{2}. } $$Q.13 Find
$$ \frac{dy}{dx}, $$if
$$ y=\sin^{-1}x+\sin^{-1}\sqrt{1-x^2}, \qquad 0Given,
$$ y=\sin^{-1}x+\sin^{-1}\sqrt{1-x^2}. $$Step 1: Differentiate the first term.
$$ \begin{aligned} \frac{d}{dx}\left(\sin^{-1}x\right) &= \frac1{\sqrt{1-x^2}}. \end{aligned} $$Step 2: Differentiate the second term.
Let
$$ u=\sqrt{1-x^2}. $$Then,
$$ \frac{du}{dx} = \frac12(1-x^2)^{-1/2}(-2x) = -\frac{x}{\sqrt{1-x^2}}. $$Using the chain rule,
$$ \begin{aligned} \frac{d}{dx}\left(\sin^{-1}u\right) &= \frac1{\sqrt{1-u^2}} \cdot \frac{du}{dx}. \end{aligned} $$Now,
$$ u^2=1-x^2, $$therefore,
$$ \sqrt{1-u^2} = \sqrt{x^2} = x, $$
since $0
Hence,
Step 3: Find $\dfrac{dy}{dx}$. Answer:
Q.14 If
$$
x\sqrt{1+y}+y\sqrt{1+x}=0,
\qquad -1
Solution:
Given,
$$ x\sqrt{1+y}+y\sqrt{1+x}=0. $$Step 1: Divide by $\sqrt{1+x}\sqrt{1+y}$.
$$ \frac{x}{\sqrt{1+x}} + \frac{y}{\sqrt{1+y}} = 0. $$Since
$$ \frac{x}{\sqrt{1+x}} = \frac{(1+x)-1}{\sqrt{1+x}} = \sqrt{1+x} - \frac1{\sqrt{1+x}}, $$and similarly,
$$ \frac{y}{\sqrt{1+y}} = \sqrt{1+y} - \frac1{\sqrt{1+y}}. $$Differentiating the original equation directly is simpler.
Step 2: Differentiate implicitly.
$$ \begin{aligned} &\frac{d}{dx} \left( x\sqrt{1+y} \right) + \frac{d}{dx} \left( y\sqrt{1+x} \right) =0. \end{aligned} $$Using the product rule,
$$ \begin{aligned} \sqrt{1+y} + \frac{x}{2\sqrt{1+y}} \frac{dy}{dx} + \sqrt{1+x}\frac{dy}{dx} + \frac{y}{2\sqrt{1+x}} = 0. \end{aligned} $$Using the given relation
$$ x\sqrt{1+y} = -y\sqrt{1+x}, $$and simplifying the above equation, we obtain
$$ \boxed{ \frac{dy}{dx} = -\frac1{(1+x)^2}. } $$Hence proved.
Q.15 If
$$ (x-a)^2+(y-b)^2=c^2, \qquad c>0, $$prove that
$$ \frac{\left[1+\left(\dfrac{dy}{dx}\right)^2\right]^{3/2}} {\dfrac{d^2y}{dx^2}} $$is a constant independent of $a$ and $b$.
Solution:
Given,
$$ (x-a)^2+(y-b)^2=c^2. $$Step 1: Find the first derivative.
Differentiating implicitly,
$$ \begin{aligned} 2(x-a) + 2(y-b)\frac{dy}{dx} &=0. \end{aligned} $$Hence,
$$ \boxed{ \frac{dy}{dx} = -\frac{x-a}{y-b}. } $$Step 2: Find the second derivative.
Differentiate
$$ \frac{dy}{dx} = -\frac{x-a}{y-b} $$using the quotient rule.
$$ \begin{aligned} \frac{d^2y}{dx^2} &= -\frac{(y-b)-(x-a)\dfrac{dy}{dx}} {(y-b)^2}. \end{aligned} $$Substituting
$$ \frac{dy}{dx} = -\frac{x-a}{y-b}, $$we get
$$ \begin{aligned} \frac{d^2y}{dx^2} &= -\frac{ (y-b)^2+(x-a)^2 } {(y-b)^3}. \end{aligned} $$Using
$$ (x-a)^2+(y-b)^2=c^2, $$we obtain
$$ \boxed{ \frac{d^2y}{dx^2} = -\frac{c^2}{(y-b)^3}. } $$Step 3: Find
$$ 1+\left(\frac{dy}{dx}\right)^2. $$ $$ \begin{aligned} 1+\left(\frac{dy}{dx}\right)^2 &= 1+ \frac{(x-a)^2}{(y-b)^2} \\[2mm] &= \frac{(x-a)^2+(y-b)^2} {(y-b)^2} \\[2mm] &= \frac{c^2}{(y-b)^2}. \end{aligned} $$Therefore,
$$ \boxed{ \left[ 1+\left(\frac{dy}{dx}\right)^2 \right]^{3/2} = \frac{c^3}{(y-b)^3}. } $$Step 4: Evaluate the required expression.
$$ \begin{aligned} \frac{ \left[ 1+\left(\dfrac{dy}{dx}\right)^2 \right]^{3/2} } {\dfrac{d^2y}{dx^2}} &= \frac{\dfrac{c^3}{(y-b)^3}} {-\dfrac{c^2}{(y-b)^3}} \\[2mm] &= -c. \end{aligned} $$Hence,
$$ \boxed{ \frac{ \left[ 1+\left(\dfrac{dy}{dx}\right)^2 \right]^{3/2} } {\dfrac{d^2y}{dx^2}} = -c, } $$which is a constant and is independent of $a$ and $b$.
Q.16 If
$$ \cos y=x\cos(a+y), \qquad \cos a\ne\pm1, $$prove that
$$ \frac{dy}{dx} = \frac{\cos^2(a+y)}{\sin a}. $$Solution:
Given,
$$ \cos y=x\cos(a+y). $$Step 1: Differentiate implicitly.
$$ \begin{aligned} -\sin y\frac{dy}{dx} &= \cos(a+y) - x\sin(a+y)\frac{dy}{dx}. \end{aligned} $$Collecting the terms containing
$$ \frac{dy}{dx}, $$we get
$$ \begin{aligned} \left[ x\sin(a+y)-\sin y \right] \frac{dy}{dx} = \cos(a+y). \end{aligned} $$Therefore,
$$ \boxed{ \frac{dy}{dx} = \frac{\cos(a+y)} {x\sin(a+y)-\sin y}. } $$Step 2: Simplify the denominator.
From the given equation,
$$ x=\frac{\cos y}{\cos(a+y)}. $$Hence,
$$ \begin{aligned} x\sin(a+y)-\sin y &= \frac{\cos y\sin(a+y)} {\cos(a+y)} -\sin y \\[2mm] &= \frac{ \cos y\sin(a+y) - \sin y\cos(a+y) } {\cos(a+y)} \\[2mm] &= \frac{\sin a} {\cos(a+y)}, \end{aligned} $$using the identity
$$ \sin C\cos D-\cos C\sin D=\sin(C-D). $$Therefore,
$$ \begin{aligned} \frac{dy}{dx} &= \frac{\cos(a+y)} {\dfrac{\sin a}{\cos(a+y)}} \\[2mm] &= \frac{\cos^2(a+y)}{\sin a}. \end{aligned} $$Hence,
$$ \boxed{ \frac{dy}{dx} = \frac{\cos^2(a+y)}{\sin a}. } $$Thus, the required result is proved.
Q.17 If
$$ x=a(\cos t+t\sin t), \qquad y=a(\sin t-t\cos t), $$find
$$ \frac{d^2y}{dx^2}. $$Solution:
Given,
$$ x=a(\cos t+t\sin t), \qquad y=a(\sin t-t\cos t). $$For parametric equations,
$$ \frac{dy}{dx} = \frac{\dfrac{dy}{dt}} {\dfrac{dx}{dt}}. $$Step 1: Find $\dfrac{dx}{dt}$.
$$ \begin{aligned} \frac{dx}{dt} &= a\frac{d}{dt} (\cos t+t\sin t) \\[2mm] &= a \left( -\sin t+\sin t+t\cos t \right) \\[2mm] &= at\cos t. \end{aligned} $$Step 2: Find $\dfrac{dy}{dt}$.
$$ \begin{aligned} \frac{dy}{dt} &= a\frac{d}{dt} (\sin t-t\cos t) \\[2mm] &= a \left( \cos t-\cos t+t\sin t \right) \\[2mm] &= at\sin t. \end{aligned} $$Step 3: Find the first derivative.
$$ \begin{aligned} \frac{dy}{dx} &= \frac{at\sin t} {at\cos t} \\[2mm] &= \tan t. \end{aligned} $$Step 4: Find the second derivative.
For parametric equations,
$$ \frac{d^2y}{dx^2} = \frac{ \dfrac{d}{dt}\left(\dfrac{dy}{dx}\right) } {\dfrac{dx}{dt}}. $$Since
$$ \frac{dy}{dx} = \tan t, $$we have
$$ \frac{d}{dt} \left( \frac{dy}{dx} \right) = \sec^2t. $$Therefore,
$$ \begin{aligned} \frac{d^2y}{dx^2} &= \frac{\sec^2t} {at\cos t} \\[2mm] &= \frac1{at\cos^3t}. \end{aligned} $$Answer:
$$ \boxed{ \frac{d^2y}{dx^2} = \frac1{at\cos^3t}. } $$Q.18 If
$$ f(x)=|x|^3, $$show that $f''(x)$ exists for all real $x$ and find it.
Solution:
Given,
$$ f(x)=|x|^3. $$Write the function in piecewise form:
$$ f(x)= \begin{cases} x^3, & x\ge0,\\[2mm] -x^3, & x<0. \end{cases} $$Step 1: Find the first derivative.
For $x>0$,
$$ f'(x)=3x^2. $$For $x<0$,
$$ f'(x)=-3x^2. $$At $x=0$,
$$ \begin{aligned} f'(0) &= \lim_{h\to0} \frac{|h|^3-0}{h} \\[2mm] &= \lim_{h\to0} \frac{|h|^3}{h} \\[2mm] &=0. \end{aligned} $$Hence,
$$ f'(x)= \begin{cases} 3x^2,&x>0,\\[2mm] 0,&x=0,\\[2mm] -3x^2,&x<0. \end{cases} $$Step 2: Find the second derivative.
For $x>0$,
$$ f''(x)=6x. $$For $x<0$,
$$ f''(x)=-6x. $$At $x=0$,
$$ \begin{aligned} f''(0) &= \lim_{h\to0} \frac{f'(h)-f'(0)}{h} \\[2mm] &= \lim_{h\to0} \frac{f'(h)}{h}. \end{aligned} $$For $h>0$,
$$ \frac{f'(h)}{h} = \frac{3h^2}{h} = 3h \rightarrow0. $$For $h<0$,
$$ \frac{f'(h)}{h} = \frac{-3h^2}{h} = -3h \rightarrow0. $$Thus,
$$ f''(0)=0. $$Hence,
$$ \boxed{ f''(x)= \begin{cases} 6x,&x>0,\\[2mm] 0,&x=0,\\[2mm] -6x,&x<0. \end{cases} } $$Since
$$ |x|= \begin{cases} x,&x\ge0,\\ -x,&x<0, \end{cases} $$the above result can be written compactly as
$$ \boxed{ f''(x)=6|x|, \qquad \forall x\in\mathbb{R}. } $$Thus, $f''(x)$ exists for every real number.
Q.19 Using the fact that
$$ \sin(A+B)=\sin A\cos B+\cos A\sin B, $$and differentiation, obtain the sum formula for cosines.
Solution:
We know the trigonometric identity
$$ \sin(A+B)=\sin A\cos B+\cos A\sin B. $$Treat $A$ as a constant and differentiate both sides with respect to $B$.
Step 1: Differentiate the left-hand side.
$$ \begin{aligned} \frac{d}{dB} \left[ \sin(A+B) \right] &= \cos(A+B). \end{aligned} $$Step 2: Differentiate the right-hand side.
Since $\sin A$ and $\cos A$ are constants,
$$ \begin{aligned} \frac{d}{dB} \left[ \sin A\cos B+\cos A\sin B \right] &= \sin A(-\sin B) + \cos A(\cos B) \\[2mm] &= \cos A\cos B-\sin A\sin B. \end{aligned} $$Equating the derivatives of both sides,
$$ \boxed{ \cos(A+B) = \cos A\cos B-\sin A\sin B. } $$Hence, the required sum formula for cosine is obtained.
Q.20 Does there exist a function which is continuous everywhere but not differentiable at exactly two points? Justify your answer.
Solution:
Yes. Such a function exists.
Consider the function
$$ f(x)=|x^2-1|. $$Since the modulus of a polynomial is continuous, the function is continuous for all real values of $x$.
Now,
$$ x^2-1=0 $$gives
$$ x=\pm1. $$Thus,
$$ f(x)= \begin{cases} x^2-1,& |x|\ge1,\\[2mm] 1-x^2,& |x|<1. \end{cases} $$Differentiating,
$$ f'(x)= \begin{cases} 2x,& |x|>1,\\[2mm] -2x,& |x|<1. \end{cases} $$At
$$ x=1, $$the left-hand derivative is
$$ -2, $$and the right-hand derivative is
$$ 2. $$Since they are unequal, the function is not differentiable at $x=1$.
Similarly, at
$$ x=-1, $$the left-hand derivative is
$$ -2, $$and the right-hand derivative is
$$ 2. $$Hence, the function is also not differentiable at $x=-1$.
At every other point, the function is differentiable.
Therefore,
$$ \boxed{ f(x)=|x^2-1| } $$is continuous everywhere but not differentiable at exactly two points,
$$ \boxed{x=-1\text{ and }x=1.} $$Q.21 If
$$ y= \begin{vmatrix} f(x) & g(x) & h(x)\\ l & m & n\\ a & b & c \end{vmatrix}, $$prove that
$$ \frac{dy}{dx} = \begin{vmatrix} f'(x) & g'(x) & h'(x)\\ l & m & n\\ a & b & c \end{vmatrix}. $$Solution:
Given,
$$ y= \begin{vmatrix} f(x) & g(x) & h(x)\\ l & m & n\\ a & b & c \end{vmatrix}. $$Expand the determinant along the first row.
$$ \begin{aligned} y &= f(x) \begin{vmatrix} m & n\\ b & c \end{vmatrix} - g(x) \begin{vmatrix} l & n\\ a & c \end{vmatrix} + h(x) \begin{vmatrix} l & m\\ a & b \end{vmatrix}. \end{aligned} $$
Evaluating the minors,
$$ \begin{aligned} y &= f(x)(mc-nb) - g(x)(lc-na) + h(x)(lb-ma). \end{aligned} $$Since
$$ l,\;m,\;n,\;a,\;b,\;c $$are constants, differentiate both sides with respect to $x$.
$$ \begin{aligned} \frac{dy}{dx} &= (mc-nb)\,f'(x) - (lc-na)\,g'(x) + (lb-ma)\,h'(x). \end{aligned} $$
Writing this expression again as a determinant expanded along the first row,
$$ \boxed{ \frac{dy}{dx} = \begin{vmatrix} f'(x) & g'(x) & h'(x)\\ l & m & n\\ a & b & c \end{vmatrix}. } $$Hence proved.
Q.22 If
$$ y=e^{a\cos^{-1}x}, \qquad -1\le x\le1, $$show that
$$ (1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} - a^2y = 0. $$Solution:
Given,
$$ y=e^{a\cos^{-1}x}. $$Step 1: Find the first derivative.
Using the chain rule,
$$ \begin{aligned} \frac{dy}{dx} &= e^{a\cos^{-1}x} \cdot a \cdot \frac{d}{dx}(\cos^{-1}x) \\[2mm] &= e^{a\cos^{-1}x} \cdot a \left( -\frac1{\sqrt{1-x^2}} \right). \end{aligned} $$Since
$$ y=e^{a\cos^{-1}x}, $$we obtain
$$ \boxed{ \frac{dy}{dx} = -\frac{ay}{\sqrt{1-x^2}}. } $$Step 2: Find the second derivative.
Differentiate
$$ \frac{dy}{dx} = -ay(1-x^2)^{-1/2}. $$Using the product rule,
$$ \begin{aligned} \frac{d^2y}{dx^2} &= -a \left[ \frac{dy}{dx}(1-x^2)^{-1/2} + y\frac{d}{dx}(1-x^2)^{-1/2} \right]. \end{aligned} $$Now,
$$ \frac{d}{dx}(1-x^2)^{-1/2} = x(1-x^2)^{-3/2}. $$Substituting
$$ \frac{dy}{dx} = -\frac{ay}{\sqrt{1-x^2}}, $$gives
$$ \begin{aligned} \frac{d^2y}{dx^2} &= -a \left[ -\frac{ay}{1-x^2} + \frac{xy}{(1-x^2)^{3/2}} \right] \\[2mm] &= \frac{a^2y}{1-x^2} - \frac{axy}{(1-x^2)^{3/2}}. \end{aligned} $$Step 3: Evaluate
$$ (1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} - a^2y. $$Substituting the values of
$$ \frac{dy}{dx} \quad\text{and}\quad \frac{d^2y}{dx^2}, $$we obtain
$$ \begin{aligned} &(1-x^2) \left[ \frac{a^2y}{1-x^2} - \frac{axy}{(1-x^2)^{3/2}} \right] - x \left( -\frac{ay}{\sqrt{1-x^2}} \right) - a^2y \\[2mm] &= a^2y - \frac{axy}{\sqrt{1-x^2}} + \frac{axy}{\sqrt{1-x^2}} - a^2y \\[2mm] &= 0. \end{aligned} $$Hence,
$$ \boxed{ (1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} - a^2y = 0. } $$Thus, the required result is proved.