Solved Examples

Important Theorems

Theorem 1 (Algebra of Continuous Functions)

Suppose $f$ and $g$ be two real functions continuous at a real number $c$. Then:

1. $f + g$ is continuous at $x = c$.
2. $f - g$ is continuous at $x = c$.
3. $f \cdot g$ is continuous at $x = c$.
4. $\left(\frac{f}{g}\right)$ is continuous at $x = c$, provided $g(c) \neq 0$.

Theorem 2 (Continuity of Composite Functions)

Suppose $f$ and $g$ are real valued functions such that $(f \circ g)$ is defined at $c$. If $g$ is continuous at $c$ and $f$ is continuous at $g(c)$, then $(f \circ g)$ is continuous at $c$.

Theorem 3 (Differentiability implies Continuity)

If a function $f$ is differentiable at a point $c$, then it is also continuous at that point. The converse is not always true.

Theorem 4 (Chain Rule)

Let $f = v \circ u$. If $t = u(x)$ and both $\frac{dt}{dx}$ and $\frac{dv}{dt}$ exist, then: $$\frac{df}{dx} = \frac{dv}{dt} \cdot \frac{dt}{dx}$$

Theorem 5 (Logarithmic Differentiation)

For a function of the form $y = u(x)^{v(x)}$, where $u(x) > 0$, we take logarithm on both sides: $$\ln y = v(x) \ln u(x)$$ and then differentiate implicitly with respect to $x$.

Example 1: Checking Continuity at a Point

Check the continuity of the function $f$ given by $f(x) = 2x + 3$ at $x = 1$.

Solution: Step 1: Find the value of the function at $x = 1$. $$f(1) = 2(1) + 3 = 5.$$

Step 2: Find the limit of the function as $x \to 1$. Since $2x+3$ is a polynomial, its limit at any point is obtained by direct substitution: $$\lim_{x \to 1} (2x+3) = 2(1)+3 = 5.$$

Step 3: Compare the two values. $$\lim_{x \to 1} f(x) = 5 = f(1).$$ Hence, the function is continuous at $x=1$.

Answer: The function is continuous at $x = 1$.

Example 2: Continuity of Modulus Function

Discuss the continuity of the function $f(x) = |x|$ at $x = 0$.

Solution: Step 1: Write $|x|$ in piecewise form: $$f(x) = \begin{cases} -x, & x<0 \\ x, & x \ge 0 \end{cases}$$

Step 2: Find the left-hand limit at $x=0$. $$\lim_{x \to 0^-} |x| = \lim_{x \to 0^-} (-x) = 0.$$

Step 3: Find the right-hand limit at $x=0$. $$\lim_{x \to 0^+} |x| = \lim_{x \to 0^+} x = 0.$$

Step 4: Find the function value. $$f(0)=|0|=0.$$

Step 5: Since $$\lim_{x \to 0^-} f(x)=\lim_{x \to 0^+} f(x)=f(0)=0,$$ $f(x)$ is continuous at $x=0$.

Answer: The function is continuous at $x = 0$.

Example 3: Finding an Unknown Constant for Continuity

Find the value of $k$ so that the function $f(x) = \begin{cases} kx + 1, & \text{if } x \le \pi \\ \cos x, & \text{if } x > \pi \end{cases}$ is continuous at $x = \pi$.

Solution: Step 1: For continuity at $x=\pi$, we must have $$\lim_{x \to \pi^-} f(x)=\lim_{x \to \pi^+} f(x)=f(\pi).$$

Step 2: Left-hand limit: $$\lim_{x \to \pi^-}(kx+1)=k\pi+1.$$ Since the first branch is also defined at $x=\pi$, $$f(\pi)=k\pi+1.$$

Step 3: Right-hand limit: $$\lim_{x \to \pi^+} \cos x = \cos \pi = -1.$$

Step 4: For continuity, $$k\pi+1=-1.$$ So, $$k\pi=-2 \implies k=-\frac{2}{\pi}.$$

Answer: $k = -\frac{2}{\pi}$

Example 4: Continuity Involving Trigonometry and Limits

Determine the value of $k$ for which the function $f(x) = \begin{cases} \frac{1 - \cos 4x}{8x^2}, & \text{if } x \ne 0 \\ k, & \text{if } x = 0 \end{cases}$ is continuous at $x = 0$.

Solution: Step 1: Continuity at $x=0$ requires $$\lim_{x \to 0} \frac{1-\cos 4x}{8x^2} = k.$$

Step 2: Use the identity $1-\cos 2\theta = 2\sin^2 \theta$ with $2\theta=4x$. Then, $$1-\cos 4x = 2\sin^2 2x.$$ So, $$\lim_{x \to 0} \frac{1-\cos 4x}{8x^2} = \lim_{x \to 0} \frac{2\sin^2 2x}{8x^2} = \lim_{x \to 0} \frac{\sin^2 2x}{4x^2}.$$

Step 3: Rewrite the expression as $$\left(\frac{\sin 2x}{2x}\right)^2.$$ Now use the standard limit $$\lim_{u \to 0} \frac{\sin u}{u}=1.$$ Hence, $$\lim_{x \to 0} \left(\frac{\sin 2x}{2x}\right)^2 = 1.$$

Step 4: Therefore, for continuity, $$k=1.$$

Answer: $k = 1$

Example 5: Checking Differentiability

Prove that the function given by $f(x) = |x - 1|, x \in \mathbb{R}$ is not differentiable at $x = 1$.

Solution: Step 1: Find the left-hand derivative at $x=1$. $$LHD = \lim_{h \to 0^-} \frac{f(1+h)-f(1)}{h}.$$ Now, $$f(1)=|1-1|=0.$$ For $h<0$, we have $|1+h-1|=|h|=-h$. Thus, $$LHD = \lim_{h \to 0^-} \frac{-h-0}{h} = -1.$$

Step 2: Find the right-hand derivative at $x=1$. $$RHD = \lim_{h \to 0^+} \frac{f(1+h)-f(1)}{h}.$$ For $h>0$, $|h|=h$, so $$RHD = \lim_{h \to 0^+} \frac{h-0}{h}=1.$$

Step 3: Since $$LHD \ne RHD,$$ $f$ is not differentiable at $x=1$.

Answer: Proved analytically.

Example 6: Differentiability of Greatest Integer Function

Show that the greatest integer function defined by $f(x) = [x]$, $0 < x < 3$ is not differentiable at $x = 1$.

Solution: Step 1: Find the left-hand derivative. $$LHD = \lim_{h \to 0^-} \frac{f(1+h)-f(1)}{h}.$$ For $h<0$, $1+h$ is slightly less than $1$, so $$[1+h]=0,$$ and $$f(1)=[1]=1.$$ Therefore, $$LHD = \lim_{h \to 0^-} \frac{0-1}{h} = \lim_{h \to 0^-} \frac{-1}{h} = +\infty.$$ So the left-hand derivative does not exist as a finite number.

Step 2: Find the right-hand derivative. For $h>0$, $1+h$ is slightly more than $1$, so $$[1+h]=1.$$ Thus, $$RHD = \lim_{h \to 0^+} \frac{1-1}{h}=0.$$

Step 3: Since the left-hand derivative is not finite and is not equal to the right-hand derivative, the function is not differentiable at $x=1$.

Answer: Proved.

Example 7: Basic Chain Rule

Differentiate $\sin(x^2 + 5)$ with respect to $x$.

Solution: Step 1: Let $$u = x^2+5.$$ Then the function becomes $$y = \sin u.$$

Step 2: Differentiate using chain rule: $$\frac{dy}{dx} = \frac{dy}{du}\cdot \frac{du}{dx} = \cos u \cdot 2x.$$

Step 3: Substitute back $u=x^2+5$. $$\frac{dy}{dx} = 2x\cos(x^2+5).$$

Answer: $2x \cos(x^2 + 5)$

Example 8: Chain Rule with Double Trigonometry

Differentiate $\cos(\sin x)$ with respect to $x$.

Solution: Step 1: Let $$u = \sin x.$$ Then, $$y = \cos u.$$

Step 2: Differentiate using chain rule: $$\frac{dy}{dx} = -\sin u \cdot \frac{du}{dx}.$$ Since $$\frac{du}{dx}=\cos x,$$ we get $$\frac{dy}{dx} = -\sin(\sin x)\cos x.$$

Answer: $-\cos x \sin(\sin x)$

Example 9: Combining Quotient and Chain Rules

Differentiate $y = \frac{\sin(ax + b)}{\cos(cx + d)}$.

Solution: Step 1: Let $$u = \sin(ax+b), \qquad v = \cos(cx+d).$$ Then $$y = \frac{u}{v}.$$

Step 2: Use the quotient rule: $$\frac{dy}{dx} = \frac{u'v-uv'}{v^2}.$$

Step 3: Differentiate $u$ and $v$ by chain rule: $$u' = a\cos(ax+b),$$ $$v' = -c\sin(cx+d).$$

Step 4: Substitute into the quotient rule: $$\frac{dy}{dx} = \frac{a\cos(ax+b)\cos(cx+d) - \sin(ax+b)(-c\sin(cx+d))}{\cos^2(cx+d)}.$$

Step 5: Simplify: $$\frac{dy}{dx} = \frac{a\cos(ax+b)\cos(cx+d) + c\sin(ax+b)\sin(cx+d)}{\cos^2(cx+d)}.$$

Answer: $\frac{a \cos(ax + b) \cos(cx + d) + c \sin(ax + b) \sin(cx + d)}{\cos^2(cx + d)}$

Example 10: Basic Implicit Differentiation

Find $\frac{dy}{dx}$ if $2x + 3y = \sin x$.

Solution: Step 1: Differentiate both sides with respect to $x$: $$\frac{d}{dx}(2x)+\frac{d}{dx}(3y)=\frac{d}{dx}(\sin x).$$

Step 2: This gives $$2 + 3\frac{dy}{dx} = \cos x.$$

Step 3: Solve for $\frac{dy}{dx}$: $$3\frac{dy}{dx} = \cos x - 2,$$ so $$\frac{dy}{dx} = \frac{\cos x - 2}{3}.$$

Answer: $\frac{\cos x - 2}{3}$

Example 11: Advanced Implicit Differentiation with Product Rule

Find $\frac{dy}{dx}$ if $\sin^2 y + \cos(xy) = \pi$.

Solution: Step 1: Differentiate both sides with respect to $x$: $$\frac{d}{dx}(\sin^2 y) + \frac{d}{dx}(\cos(xy)) = 0.$$

Step 2: Differentiate each term carefully. For $\sin^2 y$: $$\frac{d}{dx}(\sin^2 y)=2\sin y\cos y\frac{dy}{dx}=\sin 2y\frac{dy}{dx}.$$ For $\cos(xy)$: $$\frac{d}{dx}(\cos(xy)) = -\sin(xy)\cdot \frac{d}{dx}(xy).$$ Using the product rule, $$\frac{d}{dx}(xy)=x\frac{dy}{dx}+y.$$ So, $$\frac{d}{dx}(\cos(xy)) = -\sin(xy)\left(x\frac{dy}{dx}+y\right).$$

Step 3: Substitute into the differentiated equation: $$\sin 2y\frac{dy}{dx} - x\sin(xy)\frac{dy}{dx} - y\sin(xy)=0.$$

Step 4: Collect terms containing $\frac{dy}{dx}$: $$\frac{dy}{dx}[\sin 2y - x\sin(xy)] = y\sin(xy).$$

Step 5: Therefore, $$\frac{dy}{dx} = \frac{y\sin(xy)}{\sin 2y - x\sin(xy)}.$$

Answer: $\frac{y\sin(xy)}{\sin(2y) - x\sin(xy)}$

Example 12: Inverse Trigonometry Simplification

Find $\frac{dy}{dx}$ if $y = \tan^{-1} \left( \frac{\sin x}{1 + \cos x} \right)$.

Solution: Step 1: Use the half-angle identity $$\frac{\sin x}{1+\cos x} = \tan\frac{x}{2}$$ whenever the expression is defined. Thus, $$y = \tan^{-1}\left(\tan\frac{x}{2}\right).$$

Step 2: Instead of directly replacing $y$ by $x/2$ globally, differentiate using the chain rule with $$u = \tan\frac{x}{2}.$$ Then $$\frac{dy}{dx} = \frac{1}{1+u^2}\cdot \frac{du}{dx}.$$

Step 3: Compute $u'$: $$\frac{du}{dx} = \sec^2\frac{x}{2}\cdot \frac{1}{2}.$$ Since $$1+u^2 = 1+\tan^2\frac{x}{2} = \sec^2\frac{x}{2},$$ we get $$\frac{dy}{dx} = \frac{1}{\sec^2(x/2)} \cdot \frac{1}{2}\sec^2\frac{x}{2} = \frac{1}{2}.$$

Answer: $\frac{1}{2}$

Example 13: Trigonometric Substitution (Sine)

Find $\frac{dy}{dx}$ if $y = \sin^{-1} \left( \frac{2x}{1 + x^2} \right)$.

Solution: Step 1: The expression $$\frac{2x}{1+x^2}$$ suggests the substitution $$x = \tan \theta.$$ Then $$\theta = \tan^{-1}x.$$

Step 2: Substitute into the expression: $$y = \sin^{-1}\left(\frac{2\tan\theta}{1+\tan^2\theta}\right).$$ Using $$\sin 2\theta = \frac{2\tan\theta}{1+\tan^2\theta},$$ we get $$y = \sin^{-1}(\sin 2\theta).$$

Step 3: For the principal-value branch used in such standard problems, $$y = 2\theta = 2\tan^{-1}x.$$

Step 4: Differentiate: $$\frac{dy}{dx} = 2\cdot \frac{1}{1+x^2} = \frac{2}{1+x^2}.$$

Answer: $\frac{2}{1 + x^2}$

Example 14: Trigonometric Substitution (Secant)

Find $\frac{dy}{dx}$ if $y = \sec^{-1} \left( \frac{1}{2x^2 - 1} \right)$, $0 < x < \frac{1}{\sqrt{2}}$.

Solution: Step 1: Notice that $2x^2-1$ resembles the identity $$\cos 2\theta = 2\cos^2\theta - 1.$$ Let $$x = \cos\theta.$$ Then $$\theta = \cos^{-1}x.$$ Since $0<x<\frac{1}{\sqrt2}$, we have $\theta \in \left(\frac{\pi}{4},\frac{\pi}{2}\right)$, so $2\theta \in \left(\frac{\pi}{2},\pi\right)$, which lies in the principal range of $\sec^{-1}$.

Step 2: Substitute: $$y = \sec^{-1}\left(\frac{1}{2\cos^2\theta-1}\right)=\sec^{-1}\left(\frac{1}{\cos 2\theta}\right)=\sec^{-1}(\sec 2\theta).$$ Hence, $$y=2\theta=2\cos^{-1}x.$$

Step 3: Differentiate: $$\frac{dy}{dx} = 2\cdot \frac{d}{dx}(\cos^{-1}x)=2\left(-\frac{1}{\sqrt{1-x^2}}\right)=-\frac{2}{\sqrt{1-x^2}}.$$

Answer: $\frac{-2}{\sqrt{1 - x^2}}$

Example 15: Exponential Chain Rule

Differentiate $y = e^{\sin^{-1}x}$ with respect to $x$.

Solution: Step 1: Let $$u = \sin^{-1}x.$$ Then $$y=e^u.$$

Step 2: Apply the chain rule: $$\frac{dy}{dx} = e^u\cdot \frac{du}{dx}.$$ Now, $$\frac{du}{dx}=\frac{1}{\sqrt{1-x^2}}.$$

Step 3: Substitute back: $$\frac{dy}{dx} = e^{\sin^{-1}x}\cdot \frac{1}{\sqrt{1-x^2}}.$$

Answer: $\frac{e^{\sin^{-1}x}}{\sqrt{1 - x^2}}$

Example 16: Basic Logarithmic Differentiation

Differentiate $y = x^{\sin x}, x > 0$.

Solution: Step 1: Take natural logarithm on both sides: $$\ln y = \ln(x^{\sin x}) = \sin x \ln x.$$

Step 2: Differentiate implicitly: $$\frac{1}{y}\frac{dy}{dx} = \sin x \cdot \frac{1}{x} + \ln x \cdot \cos x.$$

Step 3: Multiply by $y$: $$\frac{dy}{dx} = y\left(\frac{\sin x}{x}+\cos x\ln x\right).$$

Step 4: Replace $y$ by $x^{\sin x}$: $$\frac{dy}{dx} = x^{\sin x}\left(\frac{\sin x}{x}+\cos x\ln x\right).$$

Answer: $x^{\sin x} \left[ \frac{\sin x}{x} + \cos x \ln x \right]$

Example 17: Logarithmic Differentiation with Sums

Find $\frac{dy}{dx}$ of $y = x^x - 2^{\sin x}$.

Solution: Step 1: Since the expression is a difference, do not take logarithm of the whole function. Instead write $$y=u-v,$$ where $$u=x^x, \qquad v=2^{\sin x}.$$ Then, $$\frac{dy}{dx}=\frac{du}{dx}-\frac{dv}{dx}.$$

Step 2: Differentiate $u=x^x$ using logarithmic differentiation. Take logs: $$\ln u = x\ln x.$$ Differentiate: $$\frac{1}{u}\frac{du}{dx} = 1+\ln x.$$ So, $$\frac{du}{dx}=u(1+\ln x)=x^x(1+\ln x).$$

Step 3: Differentiate $v=2^{\sin x}$. Using $$\frac{d}{dx}(a^{f(x)})=a^{f(x)}\ln a \cdot f'(x),$$ we get $$\frac{dv}{dx}=2^{\sin x}\ln 2 \cdot \cos x.$$

Step 4: Subtract: $$\frac{dy}{dx}=x^x(1+\ln x)-2^{\sin x}\cos x\ln 2.$$

Answer: $x^x(1 + \ln x) - 2^{\sin x} \cos x \ln 2$

Example 18: Implicit Equation with Variables in Exponents

Find $\frac{dy}{dx}$ if $y^x = x^y$.

Solution: Step 1: Take logarithm on both sides: $$\ln(y^x)=\ln(x^y).$$ So, $$x\ln y = y\ln x.$$

Step 2: Differentiate both sides with respect to $x$: $$\frac{d}{dx}(x\ln y)=\frac{d}{dx}(y\ln x).$$ Using the product rule, $$\ln y + x\cdot \frac{1}{y}\frac{dy}{dx} = \frac{y}{x} + \ln x\frac{dy}{dx}.$$

Step 3: Collect the $\frac{dy}{dx}$ terms: $$\frac{x}{y}\frac{dy}{dx} - \ln x\frac{dy}{dx} = \frac{y}{x} - \ln y.$$

Step 4: Factor out $\frac{dy}{dx}$: $$\frac{dy}{dx}\left(\frac{x}{y}-\ln x\right)=\frac{y}{x}-\ln y.$$

Step 5: Solve for $\frac{dy}{dx}$: $$\frac{dy}{dx} = \frac{\frac{y}{x}-\ln y}{\frac{x}{y}-\ln x}.$$ Multiplying numerator and denominator suitably, $$\frac{dy}{dx} = \frac{y(y-x\ln y)}{x(x-y\ln x)}.$$

Answer: $\frac{y(y - x\ln y)}{x(x - y\ln x)}$

Example 19: Infinite Series Logarithmic Differentiation

If $y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + ... \infty}}}$, prove that $\frac{dy}{dx} = \frac{\cos x}{2y - 1}$.

Solution: Step 1: Since the radical repeats infinitely, the entire repeated part is equal to $y$ itself. Hence, $$y=\sqrt{\sin x + y}.$$

Step 2: Square both sides: $$y^2 = \sin x + y.$$

Step 3: Differentiate implicitly: $$2y\frac{dy}{dx} = \cos x + \frac{dy}{dx}.$$

Step 4: Bring the derivative terms together: $$2y\frac{dy}{dx} - \frac{dy}{dx} = \cos x.$$

Step 5: Factor out $\frac{dy}{dx}$: $$\frac{dy}{dx}(2y-1)=\cos x.$$ Therefore, $$\frac{dy}{dx}=\frac{\cos x}{2y-1}.$$

Answer: Proved.

Example 20: Basic Parametric Differentiation

Find $\frac{dy}{dx}$, if $x = at^2$ and $y = 2at$.

Solution: Step 1: Differentiate $x$ and $y$ with respect to $t$: $$\frac{dx}{dt}=2at, \qquad \frac{dy}{dt}=2a.$$

Step 2: Use the parametric formula: $$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2a}{2at} = \frac{1}{t}.$$

Answer: $\frac{1}{t}$

Example 21: Parametric Differentiation of a Cycloid

Find $\frac{dy}{dx}$ if $x = a(\theta - \sin\theta)$ and $y = a(1 - \cos\theta)$.

Solution: Step 1: Differentiate with respect to $\theta$: $$\frac{dx}{d\theta}=a(1-\cos\theta), \qquad \frac{dy}{d\theta}=a\sin\theta.$$

Step 2: Therefore, $$\frac{dy}{dx}=\frac{a\sin\theta}{a(1-\cos\theta)} = \frac{\sin\theta}{1-\cos\theta}.$$

Step 3: Simplify using half-angle identities: $$\sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2},$$ $$1-\cos\theta = 2\sin^2\frac{\theta}{2}.$$ Hence, $$\frac{dy}{dx}=\frac{2\sin(\theta/2)\cos(\theta/2)}{2\sin^2(\theta/2)} = \cot\frac{\theta}{2}.$$

Answer: $\cot(\theta/2)$

Example 22: Parametric Functions Identity Trick

If $x = \sqrt{a^{\sin^{-1}t}}$ and $y = \sqrt{a^{\cos^{-1}t}}$, show that $\frac{dy}{dx} = -\frac{y}{x}$.

Solution: Step 1: Square both equations: $$x^2 = a^{\sin^{-1}t}, \qquad y^2 = a^{\cos^{-1}t}.$$

Step 2: Multiply them: $$x^2y^2 = a^{\sin^{-1}t + \cos^{-1}t}.$$ Using the identity $$\sin^{-1}t + \cos^{-1}t = \frac{\pi}{2},$$ we get $$x^2y^2 = a^{\pi/2},$$ which is a constant.

Step 3: Differentiate implicitly with respect to $x$: $$\frac{d}{dx}(x^2y^2)=0.$$ Using the product rule, $$2xy^2 + x^2\cdot 2y\frac{dy}{dx}=0.$$

Step 4: Solve for $\frac{dy}{dx}$: $$2x^2y\frac{dy}{dx} = -2xy^2,$$ so $$\frac{dy}{dx} = -\frac{y}{x}.$$

Answer: Proved.

Example 23: Second Order Derivative of a Polynomial

Find $\frac{d^2y}{dx^2}$ for $y = x^3 + \tan x$.

Solution: Step 1: First derivative: $$\frac{dy}{dx}=3x^2+\sec^2x.$$

Step 2: Differentiate again: $$\frac{d^2y}{dx^2}=6x+\frac{d}{dx}(\sec^2x).$$ Now, $$\frac{d}{dx}(\sec^2x)=2\sec x(\sec x\tan x)=2\sec^2x\tan x.$$ So, $$\frac{d^2y}{dx^2}=6x+2\sec^2x\tan x.$$

Answer: $6x + 2\sec^2 x \tan x$

Example 24: Second Order Derivative of Parametric Functions

Find $\frac{d^2y}{dx^2}$ if $x = a\cos\theta$ and $y = b\sin\theta$.

Solution: Step 1: First derivatives with respect to $\theta$ are $$\frac{dx}{d\theta}=-a\sin\theta, \qquad \frac{dy}{d\theta}=b\cos\theta.$$ Hence, $$\frac{dy}{dx}=\frac{b\cos\theta}{-a\sin\theta}=-\frac{b}{a}\cot\theta.$$

Step 2: Differentiate $\frac{dy}{dx}$ with respect to $\theta$: $$\frac{d}{d\theta}\left(-\frac{b}{a}\cot\theta\right)=\frac{b}{a}\csc^2\theta.$$

Step 3: Divide by $dx/d\theta$: $$\frac{d^2y}{dx^2} = \frac{\frac{b}{a}\csc^2\theta}{-a\sin\theta} = -\frac{b}{a^2}\csc^3\theta.$$

Answer: $-\frac{b}{a^2}\csc^3\theta$

Example 25: Proving a Differential Equation (Trigonometric)

If $y = 3\cos(\log x) + 4\sin(\log x)$, show that $x^2 y_2 + xy_1 + y = 0$.

Solution: Step 1: First derivative: $$y_1 = -3\sin(\log x)\cdot \frac{1}{x} + 4\cos(\log x)\cdot \frac{1}{x} = \frac{-3\sin(\log x)+4\cos(\log x)}{x}.$$

Step 2: Multiply both sides by $x$: $$xy_1 = -3\sin(\log x)+4\cos(\log x).$$

Step 3: Differentiate again: $$\frac{d}{dx}(xy_1)=\frac{d}{dx}\left[-3\sin(\log x)+4\cos(\log x)\right].$$ The left side gives $$xy_2 + y_1.$$ The right side becomes $$-3\cos(\log x)\cdot \frac{1}{x} - 4\sin(\log x)\cdot \frac{1}{x} = -\frac{1}{x}\big[3\cos(\log x)+4\sin(\log x)\big].$$ But the bracket is exactly $y$. So, $$xy_2 + y_1 = -\frac{y}{x}.$$

Step 4: Multiply by $x$: $$x^2y_2 + xy_1 + y = 0.$$

Answer: Proved.

Example 26: Proving a Differential Equation (Exponential)

If $y = Ae^{mx} + Be^{nx}$, show that $\frac{d^2y}{dx^2} - (m+n)\frac{dy}{dx} + mny = 0$.

Solution: Step 1: First derivative: $$y' = Ame^{mx} + Bne^{nx}.$$

Step 2: Second derivative: $$y'' = Am^2e^{mx} + Bn^2e^{nx}.$$

Step 3: Substitute into the required expression: $$y''-(m+n)y'+mny.$$ This becomes $$Am^2e^{mx}+Bn^2e^{nx}-(m+n)(Ame^{mx}+Bne^{nx})+mn(Ae^{mx}+Be^{nx}).$$

Step 4: Group coefficients of $e^{mx}$ and $e^{nx}$: For $e^{mx}$, coefficient is $$Am^2-Am(m+n)+Amn = Am^2-Am^2-Amn+Amn = 0.$$ For $e^{nx}$, coefficient is $$Bn^2-Bn(m+n)+Bmn = Bn^2-Bmn-Bn^2+Bmn=0.$$ Hence the whole expression is $0$.

Answer: Proved.

JEE Main Standard

Example 27: Differentiability of Piecewise Functions

Let $f(x) = [x^2] \sin(\pi x)$, where $[\cdot]$ is the greatest integer function. Discuss the differentiability of $f(x)$ at $x=1$.

Solution: Step 1: Study the function near $x=1$. If $x \to 1^-$, then $x^2<1$ and very close to $1$, so $$[x^2]=0.$$ Hence, $$f(x)=0\cdot \sin(\pi x)=0.$$ If $x \to 1^+$ and $x$ is close to $1$, then $1<x^2<2$, so $$[x^2]=1.$$ Hence, $$f(x)=\sin(\pi x).$$ Also, $$f(1)=[1]\sin \pi = 1\cdot 0=0.$$

Step 2: Left-hand derivative: For $h<0$, $f(1+h)=0$. Therefore, $$LHD = \lim_{h \to 0^-} \frac{0-0}{h}=0.$$

Step 3: Right-hand derivative: For $h>0$, $f(1+h)=\sin(\pi(1+h))$. So, $$RHD = \lim_{h \to 0^+} \frac{\sin(\pi+\pi h)-0}{h} = \lim_{h \to 0^+} \frac{-\sin(\pi h)}{h}.$$ Multiply and divide by $\pi$: $$RHD = -\pi \lim_{h \to 0^+} \frac{\sin(\pi h)}{\pi h} = -\pi.$$

Step 4: Since $$LHD \ne RHD,$$ $f(x)$ is not differentiable at $x=1$.

Answer: $f(x)$ is not differentiable at $x=1$.

Example 28: Differentiating a Function with Respect to Another Function

Differentiate $u(x) = \tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ with respect to $v(x) = \tan^{-1} x$.

Solution: Step 1: Let $$x = \tan\theta,$$ so that $$\theta = \tan^{-1}x = v.$$ Also, $$\sqrt{1+x^2}=\sqrt{1+\tan^2\theta}=\sec\theta.$$ Thus, $$u = \tan^{-1}\left(\frac{\sec\theta-1}{\tan\theta}\right).$$

Step 2: Simplify the inside expression: $$\frac{\sec\theta-1}{\tan\theta} = \frac{\frac{1}{\cos\theta}-1}{\frac{\sin\theta}{\cos\theta}} = \frac{1-\cos\theta}{\sin\theta}.$$ Using half-angle identities, $$\frac{1-\cos\theta}{\sin\theta} = \tan\frac{\theta}{2}.$$ Hence, $$u = \tan^{-1}\left(\tan\frac{\theta}{2}\right)=\frac{\theta}{2}$$ for the principal-value branch.

Step 3: Since $v=\theta$, we have $$u = \frac{v}{2}.$$ Therefore, $$\frac{du}{dv} = \frac{1}{2}.$$

Answer: $\frac{1}{2}$