Complete solutions for Class 12 Maths Miscellaneous Exercise on Chapter 5 (NCERT). Tackle advanced problems in Logarithmic, Implicit, and Parametric Differentiation, including derivatives of $f(x)^{g(x)}$ forms, inverse trigonometric functions, and complex proofs involving the second derivative.
This exercise covers a wide range of differentiation techniques, including the Chain Rule, Product Rule, Quotient Rule, Logarithmic Differentiation, Implicit Differentiation, and Parametric Differentiation, as well as proving identities involving derivatives.
✍️ Differentiation Solutions (Exercises 1 – 11)
1. $y = (3x^2 – 9x + 5)^9$
Use the Chain Rule:
$$\frac{dy}{dx} = 9 (3x^2 – 9x + 5)^{9-1} \cdot \frac{d}{dx}(3x^2 – 9x + 5)$$
$$\frac{dy}{dx} = 9 (3x^2 – 9x + 5)^8 \cdot (6x – 9)$$
$$\frac{dy}{dx} = 9 \cdot 3 (2x – 3) (3x^2 – 9x + 5)^8$$
$$\mathbf{\frac{dy}{dx} = 27 (2x – 3) (3x^2 – 9x + 5)^8}$$
2. $y = \sin^3 x + \cos^6 x$
Differentiate term by term using the Chain Rule:
$$\frac{dy}{dx} = \frac{d}{dx}(\sin x)^3 + \frac{d}{dx}(\cos x)^6$$
$$\frac{dy}{dx} = 3 \sin^2 x \cdot \frac{d}{dx}(\sin x) + 6 \cos^5 x \cdot \frac{d}{dx}(\cos x)$$
$$\frac{dy}{dx} = 3 \sin^2 x \cos x + 6 \cos^5 x (-\sin x)$$
$$\frac{dy}{dx} = 3 \sin x \cos x (\sin x – 2 \cos^4 x)$$
$$\mathbf{\frac{dy}{dx} = 3 \sin x \cos x (\sin x – 2 \cos^4 x)}$$
3. $y = (5x)^{3 \cos 2x}$
This is of the form $f(x)^{g(x)}$, so use Logarithmic Differentiation:
$$\log y = \log ((5x)^{3 \cos 2x})$$
$$\log y = 3 \cos 2x \cdot \log (5x)$$
Differentiate both sides w.r.t. $x$ (using the Product Rule on RHS):
$$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(3 \cos 2x) \cdot \log (5x) + 3 \cos 2x \cdot \frac{d}{dx}(\log (5x))$$
$$\frac{1}{y} \frac{dy}{dx} = (3 \cdot (-\sin 2x) \cdot 2) \log (5x) + 3 \cos 2x \cdot \left( \frac{1}{5x} \cdot 5 \right)$$
$$\frac{1}{y} \frac{dy}{dx} = -6 \sin 2x \log (5x) + \frac{3 \cos 2x}{x}$$
$$\frac{dy}{dx} = y \left[ \frac{3 \cos 2x}{x} – 6 \sin 2x \log (5x) \right]$$
$$\mathbf{\frac{dy}{dx} = (5x)^{3 \cos 2x} \left[ \frac{3 \cos 2x}{x} – 6 \sin 2x \log (5x) \right]}$$
4. $y = \sin^{-1} (x \sqrt{x}), 0 \le x \le 1$
Rewrite $x \sqrt{x}$ as $x^{1} x^{1/2} = x^{3/2}$. Use the Chain Rule:
$$y = \sin^{-1} (x^{3/2})$$
$$\frac{dy}{dx} = \frac{1}{\sqrt{1 – (x^{3/2})^2}} \cdot \frac{d}{dx}(x^{3/2})$$
$$\frac{dy}{dx} = \frac{1}{\sqrt{1 – x^3}} \cdot \frac{3}{2} x^{1/2}$$
$$\mathbf{\frac{dy}{dx} = \frac{3 \sqrt{x}}{2 \sqrt{1 – x^3}}}$$
5. $y = \frac{\cos^{-1} (x/2)}{\sqrt{2x + 7}}, -2 < x < 2$
Use the Quotient Rule: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v u’ – u v’}{v^2}$.
Let $u = \cos^{-1} (x/2)$ and $v = \sqrt{2x + 7}$.
- Find $u’$:$$u’ = \frac{d}{dx}\left(\cos^{-1} \frac{x}{2}\right) = -\frac{1}{\sqrt{1 – (x/2)^2}} \cdot \frac{1}{2} = -\frac{1}{\sqrt{1 – x^2/4}} \cdot \frac{1}{2} = -\frac{1}{\sqrt{\frac{4 – x^2}{4}}} \cdot \frac{1}{2} = -\frac{1}{\frac{\sqrt{4 – x^2}}{2}} \cdot \frac{1}{2} = -\frac{1}{\sqrt{4 – x^2}}$$
- Find $v’$:$$v’ = \frac{d}{dx}(\sqrt{2x + 7}) = \frac{1}{2\sqrt{2x + 7}} \cdot 2 = \frac{1}{\sqrt{2x + 7}}$$
- Apply the Quotient Rule:$$\frac{dy}{dx} = \frac{\sqrt{2x + 7} \left(-\frac{1}{\sqrt{4 – x^2}}\right) – \cos^{-1} (x/2) \left(\frac{1}{\sqrt{2x + 7}}\right)}{(\sqrt{2x + 7})^2}$$$$\frac{dy}{dx} = \frac{-\frac{\sqrt{2x + 7}}{\sqrt{4 – x^2}} – \frac{\cos^{-1} (x/2)}{\sqrt{2x + 7}}}{2x + 7}$$Multiply numerator and denominator by $\sqrt{4 – x^2}\sqrt{2x + 7}$:$$\frac{dy}{dx} = \frac{-(2x + 7) – \cos^{-1}(x/2) \sqrt{4 – x^2}}{(2x + 7)^{3/2} \sqrt{4 – x^2}}$$$$\mathbf{\frac{dy}{dx} = – \frac{(2x + 7) + \sqrt{4 – x^2} \cos^{-1} (x/2)}{(2x + 7)^{3/2} \sqrt{4 – x^2}}}$$
6. $y = \cot^{-1} \left[ \frac{\sqrt{1 + \sin x} + \sqrt{1 – \sin x}}{\sqrt{1 + \sin x} – \sqrt{1 – \sin x}} \right], 0 < x < \frac{\pi}{2}$
First, simplify the inner expression using trigonometry:
Since $0 < x < \frac{\pi}{2}$, we have $0 < \frac{x}{2} < \frac{\pi}{4}$. Thus $\cos \frac{x}{2} > \sin \frac{x}{2}$.
$$1 + \sin x = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2} = \left(\cos \frac{x}{2} + \sin \frac{x}{2}\right)^2$$
$$1 – \sin x = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} – 2 \sin \frac{x}{2} \cos \frac{x}{2} = \left(\cos \frac{x}{2} – \sin \frac{x}{2}\right)^2$$
$$\sqrt{1 + \sin x} = \cos \frac{x}{2} + \sin \frac{x}{2}$$
$$\sqrt{1 – \sin x} = \cos \frac{x}{2} – \sin \frac{x}{2}$$
Substitute these back into the expression:
$$\frac{\sqrt{1 + \sin x} + \sqrt{1 – \sin x}}{\sqrt{1 + \sin x} – \sqrt{1 – \sin x}} = \frac{(\cos \frac{x}{2} + \sin \frac{x}{2}) + (\cos \frac{x}{2} – \sin \frac{x}{2})}{(\cos \frac{x}{2} + \sin \frac{x}{2}) – (\cos \frac{x}{2} – \sin \frac{x}{2})}$$
$$= \frac{2 \cos \frac{x}{2}}{2 \sin \frac{x}{2}} = \cot \frac{x}{2}$$
So the function simplifies to:
$$y = \cot^{-1} \left( \cot \frac{x}{2} \right)$$
Since $0 < \frac{x}{2} < \frac{\pi}{4}$, $y = \frac{x}{2}$.
Now differentiate:
$$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{x}{2}\right)$$
$$\mathbf{\frac{dy}{dx} = \frac{1}{2}}$$
7. $y = (\log x)^{\log x}, x > 1$
Use Logarithmic Differentiation:
$$\log y = \log [(\log x)^{\log x}]$$
$$\log y = \log x \cdot \log (\log x)$$
Differentiate w.r.t. $x$ (using the Product Rule on RHS):
$$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\log x) \cdot \log (\log x) + \log x \cdot \frac{d}{dx}(\log (\log x))$$
$$\frac{1}{y} \frac{dy}{dx} = \frac{1}{x} \log (\log x) + \log x \cdot \left( \frac{1}{\log x} \cdot \frac{1}{x} \right)$$
$$\frac{1}{y} \frac{dy}{dx} = \frac{1}{x} \log (\log x) + \frac{1}{x} = \frac{1}{x} [1 + \log (\log x)]$$
$$\frac{dy}{dx} = y \cdot \frac{1}{x} [1 + \log (\log x)]$$
$$\mathbf{\frac{dy}{dx} = (\log x)^{\log x} \frac{1}{x} [1 + \log (\log x)]}$$
8. $y = \cos (a \cos x + b \sin x)$
Use the Chain Rule:
$$\frac{dy}{dx} = -\sin (a \cos x + b \sin x) \cdot \frac{d}{dx}(a \cos x + b \sin x)$$
$$\frac{dy}{dx} = -\sin (a \cos x + b \sin x) \cdot (-a \sin x + b \cos x)$$
$$\mathbf{\frac{dy}{dx} = (a \sin x – b \cos x) \sin (a \cos x + b \sin x)}$$
9. $y = (\sin x – \cos x)^{(\sin x – \cos x)}, \frac{\pi}{4} < x < \frac{3\pi}{4}$
This is of the form $f(x)^{f(x)}$, so use Logarithmic Differentiation.
Let $u = \sin x – \cos x$. Then $y = u^u$.
$$\log y = \log (u^u) = u \log u$$
Differentiate w.r.t. $x$:
$$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(u) \cdot \log u + u \cdot \frac{1}{u} \cdot \frac{d}{dx}(u)$$
$$\frac{1}{y} \frac{dy}{dx} = \frac{du}{dx} \cdot \log u + \frac{du}{dx} = \frac{du}{dx} (1 + \log u)$$
Now find $\frac{du}{dx}$:
$$\frac{du}{dx} = \frac{d}{dx}(\sin x – \cos x) = \cos x – (-\sin x) = \cos x + \sin x$$
Substitute back for $u$ and $\frac{du}{dx}$:
$$\frac{dy}{dx} = y \cdot (\cos x + \sin x) [1 + \log (\sin x – \cos x)]$$
$$\mathbf{\frac{dy}{dx} = (\sin x – \cos x)^{(\sin x – \cos x)} (\cos x + \sin x) [1 + \log (\sin x – \cos x)]}$$
10. $y = x^x + x^a + a^x + a^a$
This is the sum of four functions. Let $u = x^x$.
$$\frac{dy}{dx} = \frac{d}{dx}(x^x) + \frac{d}{dx}(x^a) + \frac{d}{dx}(a^x) + \frac{d}{dx}(a^a)$$
- $\frac{d}{dx}(x^x)$: Use logarithmic differentiation (from Ex 5.5, Q4):$$\frac{d}{dx}(x^x) = x^x (1 + \log x)$$
- $\frac{d}{dx}(x^a)$: Power Rule (since $a$ is a constant):$$\frac{d}{dx}(x^a) = a x^{a-1}$$
- $\frac{d}{dx}(a^x)$: Exponential function rule:$$\frac{d}{dx}(a^x) = a^x \log a$$
- $\frac{d}{dx}(a^a)$: Derivative of a constant:$$\frac{d}{dx}(a^a) = 0$$
Combine the results:
$$\mathbf{\frac{dy}{dx} = x^x (1 + \log x) + a x^{a-1} + a^x \log a}$$
11. $y = \frac{x^2 – 3}{\sqrt{x^2 – 3}} + \frac{x^2 + 3}{\sqrt{x^2 + 3}}$
Wait, the question is $y = \frac{x^2 – 3}{\sqrt{x^2 – 3}} + \frac{x^2 + 3}{\sqrt{x^2 + 3}}$. This can be simplified.
Assuming the intended question is $y = \frac{x^2 – 3}{\sqrt{x^2 – 3}} + \frac{x^2 + 3}{\sqrt{x^2 + 3}}$:
For $x>3$, $x^2-3 > 0$ and $x^2+3 > 0$.
$$y = \sqrt{x^2 – 3} + \sqrt{x^2 + 3}$$
Differentiate w.r.t. $x$:
$$\frac{dy}{dx} = \frac{d}{dx}(\sqrt{x^2 – 3}) + \frac{d}{dx}(\sqrt{x^2 + 3})$$
$$\frac{dy}{dx} = \frac{1}{2\sqrt{x^2 – 3}} \cdot (2x) + \frac{1}{2\sqrt{x^2 + 3}} \cdot (2x)$$
$$\mathbf{\frac{dy}{dx} = \frac{x}{\sqrt{x^2 – 3}} + \frac{x}{\sqrt{x^2 + 3}}}$$
🔁 Parametric and Implicit Differentiation (Exercises 12 – 17)
12. Find $\frac{dy}{dx}$, if $y = 12 (1 – \cos t)$, $x = 10 (t – \sin t)$
Use Parametric Differentiation: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.
- Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$:$$\frac{dx}{dt} = 10 (1 – \cos t)$$$$\frac{dy}{dt} = 12 (0 – (-\sin t)) = 12 \sin t$$
- Find $\frac{dy}{dx}$:$$\frac{dy}{dx} = \frac{12 \sin t}{10 (1 – \cos t)} = \frac{6 \sin t}{5 (1 – \cos t)}$$Use half-angle identities: $\sin t = 2 \sin \frac{t}{2} \cos \frac{t}{2}$ and $1 – \cos t = 2 \sin^2 \frac{t}{2}$.$$\frac{dy}{dx} = \frac{6 (2 \sin \frac{t}{2} \cos \frac{t}{2})}{5 (2 \sin^2 \frac{t}{2})} = \frac{6}{5} \frac{\cos \frac{t}{2}}{\sin \frac{t}{2}}$$$$\mathbf{\frac{dy}{dx} = \frac{6}{5} \cot \frac{t}{2}}$$
13. Find $\frac{dy}{dx}$, if $y = \sin^{-1} x + \sin^{-1} \sqrt{1-x^2}, 0 < x < 1$
Simplify the expression first.
Let $x = \cos \theta$, where $0 < \theta < \frac{\pi}{2}$ since $0 < x < 1$.
$$\sqrt{1 – x^2} = \sqrt{1 – \cos^2 \theta} = \sqrt{\sin^2 \theta} = \sin \theta$$
$$y = \sin^{-1} (\cos \theta) + \sin^{-1} (\sin \theta)$$
Since $0 < \theta < \frac{\pi}{2}$, $\sin^{-1}(\sin \theta) = \theta$.
Use $\cos \theta = \sin (\frac{\pi}{2} – \theta)$:
$$y = \sin^{-1} \left( \sin \left(\frac{\pi}{2} – \theta\right) \right) + \theta$$
Since $0 < \theta < \frac{\pi}{2}$, $0 < \frac{\pi}{2} – \theta < \frac{\pi}{2}$, so $\sin^{-1}(\sin A) = A$:
$$y = \left(\frac{\pi}{2} – \theta\right) + \theta = \frac{\pi}{2}$$
Differentiate the simplified form:
$$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2}\right)$$
$$\mathbf{\frac{dy}{dx} = 0}$$
14. If $x \sqrt{1+y} + y \sqrt{1+x} = 0$, for $-1 < x < 1$, prove that $\frac{dy}{dx} = -\frac{1}{(1+x)^2}$
- Solve for $y$ (by squaring, or by using substitution/tricks):$$x \sqrt{1+y} = -y \sqrt{1+x}$$Square both sides:$$x^2 (1+y) = y^2 (1+x)$$$$x^2 + x^2 y = y^2 + y^2 x$$$$x^2 – y^2 = y^2 x – x^2 y$$$$(x – y)(x + y) = xy (y – x)$$$$(x – y)(x + y) = -xy (x – y)$$Since $x \ne y$ (otherwise the original equation implies $x \sqrt{1+x} = -x \sqrt{1+x}$, which only holds if $x=0$), we can divide by $(x-y)$:$$x + y = -xy$$$$x = -y – xy = -y (1 + x)$$$$y = -\frac{x}{1 + x}$$
- Differentiate using the Quotient Rule:$$\frac{dy}{dx} = -\frac{(1 + x) \cdot 1 – x \cdot 1}{(1 + x)^2}$$$$\frac{dy}{dx} = -\frac{1 + x – x}{(1 + x)^2} = -\frac{1}{(1 + x)^2}$$$$\mathbf{\frac{dy}{dx} = -\frac{1}{(1+x)^2}}$$Hence Proved.
15. If $(x – a)^2 + (y – b)^2 = c^2$, for some $c > 0$, prove that $\frac{[1 + (dy/dx)^2]^{3/2}}{d^2 y / dx^2}$ is a constant independent of $a$ and $b$.
The expression represents the radius of curvature $\rho$.
- Find $\frac{dy}{dx}$ (Implicit Differentiation):$$2(x – a) + 2(y – b) \frac{dy}{dx} = 0$$$$(y – b) \frac{dy}{dx} = -(x – a)$$$$\frac{dy}{dx} = -\frac{x – a}{y – b}$$ (Equation 1)
- Find $\frac{d^2 y}{dx^2}$ (Quotient Rule on Eq. 1):$$\frac{d^2 y}{dx^2} = – \frac{(y – b) \cdot 1 – (x – a) \cdot \frac{dy}{dx}}{(y – b)^2}$$Substitute $\frac{dy}{dx}$ from Eq. 1 into the numerator:$$\frac{d^2 y}{dx^2} = – \frac{(y – b) – (x – a) \left(-\frac{x – a}{y – b}\right)}{(y – b)^2}$$$$\frac{d^2 y}{dx^2} = – \frac{(y – b) + \frac{(x – a)^2}{y – b}}{(y – b)^2} = – \frac{\frac{(y – b)^2 + (x – a)^2}{y – b}}{(y – b)^2}$$From the original equation, $(x – a)^2 + (y – b)^2 = c^2$.$$\frac{d^2 y}{dx^2} = – \frac{c^2}{(y – b)^3}$$ (Equation 2)
- Evaluate the expression:$$\left[1 + \left(\frac{dy}{dx}\right)^2\right] = 1 + \left(-\frac{x – a}{y – b}\right)^2 = 1 + \frac{(x – a)^2}{(y – b)^2} = \frac{(y – b)^2 + (x – a)^2}{(y – b)^2} = \frac{c^2}{(y – b)^2}$$Now substitute into the required expression:$$\frac{[1 + (dy/dx)^2]^{3/2}}{d^2 y / dx^2} = \frac{\left[\frac{c^2}{(y – b)^2}\right]^{3/2}}{-\frac{c^2}{(y – b)^3}}$$$$= \frac{\frac{(c^2)^{3/2}}{((y – b)^2)^{3/2}}}{-\frac{c^2}{(y – b)^3}} = \frac{\frac{c^3}{(y – b)^3}}{-\frac{c^2}{(y – b)^3}} = -\frac{c^3}{c^2} = -c$$Since $c$ is a constant, $\mathbf{-c}$ is a constant independent of $a$ and $b$.Hence Proved.
16. If $\cos y = x \cos (a + y)$, with $\cos a \ne \pm 1$, prove that $\frac{dy}{dx} = \frac{\cos^2 (a + y)}{\sin a}$
- Solve for $x$:$$x = \frac{\cos y}{\cos (a + y)}$$
- Differentiate w.r.t. $y$ (instead of $x$):$$\frac{dx}{dy} = \frac{\cos (a + y) \cdot \frac{d}{dy}(\cos y) – \cos y \cdot \frac{d}{dy}(\cos (a + y))}{\cos^2 (a + y)}$$$$\frac{dx}{dy} = \frac{\cos (a + y) (-\sin y) – \cos y (-\sin (a + y))}{\cos^2 (a + y)}$$$$\frac{dx}{dy} = \frac{\cos y \sin (a + y) – \sin y \cos (a + y)}{\cos^2 (a + y)}$$Use the sine addition formula: $\sin(A – B) = \sin A \cos B – \cos A \sin B$.Here, $A = a + y$ and $B = y$. $\sin((a + y) – y) = \sin a$.$$\frac{dx}{dy} = \frac{\sin ((a + y) – y)}{\cos^2 (a + y)} = \frac{\sin a}{\cos^2 (a + y)}$$
- Find $\frac{dy}{dx}$:$$\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{1}{\frac{\sin a}{\cos^2 (a + y)}} = \frac{\cos^2 (a + y)}{\sin a}$$$$\mathbf{\frac{dy}{dx} = \frac{\cos^2 (a + y)}{\sin a}}$$Hence Proved.
17. If $x = a (\cos t + t \sin t)$ and $y = a (\sin t – t \cos t)$, find $\frac{d^2 y}{dx^2}$.
- Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$ (First Derivative):$$\frac{dx}{dt} = a \left[ -\sin t + (1 \cdot \sin t + t \cdot \cos t) \right] = a t \cos t$$$$\frac{dy}{dt} = a \left[ \cos t – (1 \cdot \cos t + t \cdot (-\sin t)) \right] = a t \sin t$$
- Find $\frac{dy}{dx}$ (First Order Derivative):$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a t \sin t}{a t \cos t} = \tan t$$
- Find $\frac{d^2 y}{dx^2}$ (Second Order Derivative):$$\frac{d^2 y}{dx^2} = \frac{d}{dx} \left( \frac{dy}{dx} \right) = \frac{d}{dx} (\tan t)$$Remember to use the Chain Rule for parametric equations:$$\frac{d}{dx} (\tan t) = \frac{d}{dt} (\tan t) \cdot \frac{dt}{dx}$$$$\frac{d}{dx} (\tan t) = \sec^2 t \cdot \frac{1}{dx/dt}$$$$\frac{d^2 y}{dx^2} = \sec^2 t \cdot \frac{1}{a t \cos t} = \frac{1}{\cos^2 t} \cdot \frac{1}{a t \cos t}$$$$\mathbf{\frac{d^2 y}{dx^2} = \frac{1}{at \cos^3 t}}$$
🔬 Advanced Topics (Exercises 18 – 22)
18. If $f(x) = |x|^3$, show that $f”(x)$ exists for all real $x$ and find it.
The function is defined piecewise:
$$f(x) = \begin{cases} x^3 & \text{if } x \ge 0 \\ (-x)^3 = -x^3 & \text{if } x < 0 \end{cases}$$
- First Derivative ($f'(x)$):$$f'(x) = \begin{cases} 3x^2 & \text{if } x > 0 \\ -3x^2 & \text{if } x < 0 \end{cases}$$At $x=0$:$$f'(0)_{\text{LHL}} = \lim_{h \to 0^-} \frac{f(0+h) – f(0)}{h} = \lim_{h \to 0^-} \frac{-h^3 – 0}{h} = \lim_{h \to 0^-} -h^2 = 0$$$$f'(0)_{\text{RHL}} = \lim_{h \to 0^+} \frac{f(0+h) – f(0)}{h} = \lim_{h \to 0^+} \frac{h^3 – 0}{h} = \lim_{h \to 0^+} h^2 = 0$$Since $f'(0)_{\text{LHL}} = f'(0)_{\text{RHL}} = 0$, $f'(x)$ exists at $x=0$.We can write $f'(x) = 3x^2$ for $x \ge 0$ and $f'(x) = -3x^2$ for $x < 0$, which is equivalent to $f'(x) = 3x |x|$.
- Second Derivative ($f”(x)$):$$f”(x) = \begin{cases} 6x & \text{if } x > 0 \\ -6x & \text{if } x < 0 \end{cases}$$At $x=0$:$$f”(0)_{\text{LHL}} = \lim_{h \to 0^-} \frac{f'(0+h) – f'(0)}{h} = \lim_{h \to 0^-} \frac{-3h^2 – 0}{h} = \lim_{h \to 0^-} -3h = 0$$$$f”(0)_{\text{RHL}} = \lim_{h \to 0^+} \frac{f'(0+h) – f'(0)}{h} = \lim_{h \to 0^+} \frac{3h^2 – 0}{h} = \lim_{h \to 0^+} 3h = 0$$Since $f”(0)_{\text{LHL}} = f”(0)_{\text{RHL}} = 0$, $f”(x)$ exists at $x=0$.$$f”(x) = \begin{cases} 6x & \text{if } x \ge 0 \\ -6x & \text{if } x < 0 \end{cases}$$$$f”(x) = 6|x|$$Conclusion: $f”(x)$ exists for all real $x$.$$\mathbf{f”(x) = 6|x|}$$
19. Using the fact that $\sin (A + B) = \sin A \cos B + \cos A \sin B$ and differentiation, obtain the sum formula for cosines.
Differentiate the given identity w.r.t. $A$ (treating $B$ as a constant):
$$\frac{d}{dA} [\sin (A + B)] = \frac{d}{dA} [\sin A \cos B + \cos A \sin B]$$
$$\cos (A + B) \cdot \frac{d}{dA}(A + B) = \frac{d}{dA}(\sin A) \cos B + \sin A \frac{d}{dA}(\cos B) + \frac{d}{dA}(\cos A) \sin B + \cos A \frac{d}{dA}(\sin B)$$
$$\cos (A + B) \cdot 1 = (\cos A) \cos B + (\sin A) \cdot 0 + (-\sin A) \sin B + (\cos A) \cdot 0$$
$$\mathbf{\cos (A + B) = \cos A \cos B – \sin A \sin B}$$
This is the sum formula for cosines.
20. Does there exist a function which is continuous everywhere but not differentiable at exactly two points? Justify your answer.
Yes, such a function exists.
A function is typically not differentiable at a point where it has a sharp corner or cusp. The absolute value function $y = |x|$ is the classic example, which is continuous everywhere but not differentiable at $x=0$.
A function with exactly two sharp corners will meet the criteria. For instance:
$$f(x) = |x – 1| + |x + 1|$$
This function is the sum of two continuous functions, so it is continuous everywhere. The points where the terms change their definition (the sharp corners) are:
- $x – 1 = 0 \implies x = 1$
- $x + 1 = 0 \implies x = -1$
Thus, $f(x)$ is continuous everywhere but not differentiable at $x = -1$ and $x = 1$.
21. If $y = \begin{vmatrix} f(x) & g(x) & h(x) \\ l & m & n \\ a & b & c \end{vmatrix}$, prove that $\frac{dy}{dx} = \begin{vmatrix} f'(x) & g'(x) & h'(x) \\ l & m & n \\ a & b & c \end{vmatrix}$
The derivative of a determinant whose entries are functions of $x$ is the sum of the determinants obtained by differentiating one row (or column) at a time.
For a $3 \times 3$ determinant $y$ with function entries only in the first row, expanding along the first row:
$$y = f(x) (mc – nb) – g(x) (lc – na) + h(x) (lb – ma)$$
Note that the cofactors $(mc – nb)$, $-(lc – na)$, and $(lb – ma)$ are constants since $l, m, n, a, b, c$ are constants.
Differentiate $y$ w.r.t. $x$:
$$\frac{dy}{dx} = \frac{d}{dx}[f(x) (mc – nb)] – \frac{d}{dx}[g(x) (lc – na)] + \frac{d}{dx}[h(x) (lb – ma)]$$
$$\frac{dy}{dx} = f'(x) (mc – nb) – g'(x) (lc – na) + h'(x) (lb – ma)$$
This resulting expression is the expansion of the determinant:
$$\begin{vmatrix} f'(x) & g'(x) & h'(x) \\ l & m & n \\ a & b & c \end{vmatrix} = f'(x) (mc – nb) – g'(x) (lc – na) + h'(x) (lb – ma)$$
Thus,
$$\mathbf{\frac{dy}{dx} = \begin{vmatrix} f'(x) & g'(x) & h'(x) \\ l & m & n \\ a & b & c \end{vmatrix}}$$
Hence Proved.
22. If $y = e^{a \cos^{-1} x}$, show that $(1 – x^2) \frac{d^2 y}{dx^2} – x \frac{dy}{dx} – a^2 y = 0$.
- First Derivative ($\frac{dy}{dx}$):$$\frac{dy}{dx} = e^{a \cos^{-1} x} \cdot \frac{d}{dx}(a \cos^{-1} x)$$$$\frac{dy}{dx} = e^{a \cos^{-1} x} \cdot a \left(-\frac{1}{\sqrt{1 – x^2}}\right)$$Since $y = e^{a \cos^{-1} x}$, substitute $y$:$$\frac{dy}{dx} = -\frac{ay}{\sqrt{1 – x^2}}$$Rearrange to clear the fraction (to simplify the second derivative):$$\sqrt{1 – x^2} \frac{dy}{dx} = -ay$$ (Equation C)
- Second Derivative ($\frac{d^2 y}{dx^2}$): Differentiate Equation C w.r.t. $x$ (using the Product Rule on LHS and Implicit/Chain Rule on RHS):$$\frac{d}{dx}\left(\sqrt{1 – x^2}\right) \frac{dy}{dx} + \sqrt{1 – x^2} \frac{d}{dx}\left(\frac{dy}{dx}\right) = -a \frac{dy}{dx}$$$$\left(\frac{1}{2\sqrt{1 – x^2}} \cdot (-2x)\right) \frac{dy}{dx} + \sqrt{1 – x^2} \frac{d^2 y}{dx^2} = -a \frac{dy}{dx}$$$$-\frac{x}{\sqrt{1 – x^2}} \frac{dy}{dx} + \sqrt{1 – x^2} \frac{d^2 y}{dx^2} = -a \frac{dy}{dx}$$Multiply the entire equation by $\sqrt{1 – x^2}$ to clear the remaining fractions:$$-x \frac{dy}{dx} + (1 – x^2) \frac{d^2 y}{dx^2} = -a \sqrt{1 – x^2} \frac{dy}{dx}$$
- Final Substitution and Rearrangement:Substitute $-a \sqrt{1 – x^2} \frac{dy}{dx}$ from the rearranged Equation C, $\sqrt{1 – x^2} \frac{dy}{dx} = -ay$:$$-a \sqrt{1 – x^2} \frac{dy}{dx} = -a (-ay) = a^2 y$$Substituting $a^2 y$ into the second derivative equation:$$-x \frac{dy}{dx} + (1 – x^2) \frac{d^2 y}{dx^2} = a^2 y$$Rearrange to match the required form:$$\mathbf{(1 – x^2) \frac{d^2 y}{dx^2} – x \frac{dy}{dx} – a^2 y = 0}$$Hence Proved.