Differentiation (VSAQs)

Maths-1B | 9. Differentiation – VSAQs:
Welcome to VSAQs in Chapter 9: Differentiation. This page contains the most important VSAQs in this chapter. Aim to secure top marks in your exams by understanding these clear and straightforward Very Short Answer Questions.

---

VSAQ-1 : Find the derivative of the function f(x) = (x² – 3) (4x³ + 1)

$$f(x) = (x^2 – 3)(4x^3 + 1)$$

$$f'(x) = \frac{d}{dx}[(x^2 – 3)(4x^3 + 1)]$$

$$f'(x) = (x^2 – 3) \frac{d}{dx}(4x^3 + 1) + (4x^3 + 1) \frac{d}{dx}(x^2 – 3)$$

$$\frac{d}{dx}(4x^3 + 1) = 12x^2, \quad \frac{d}{dx}(x^2 – 3) = 2x$$

$$f'(x) = (x^2 – 3)(12x^2) + (4x^3 + 1)(2x)$$

$$f'(x) = 12x^4 – 36x^2 + 8x^4 + 2x$$

$$f'(x) = 20x^4 – 36x^2 + 2x$$

---

VSAQ-2 : If y = e^2x.log(3x + 4) then find dy/dx

$$y = e^{2x} \log(3x + 4)$$

$$\frac{dy}{dx} = e^{2x} \frac{d}{dx}[\log(3x + 4)] + \log(3x + 4) \frac{d}{dx}[e^{2x}]$$

$$\frac{d}{dx}[\log(3x + 4)] = \frac{1}{3x + 4} \cdot 3 = \frac{3}{3x + 4}$$

$$\frac{d}{dx}[e^{2x}] = e^{2x} \cdot 2$$

$$\frac{dy}{dx} = e^{2x} \left(\frac{3}{3x + 4}\right) + \log(3x + 4) \cdot 2e^{2x}$$

$$\frac{dy}{dx} = e^{2x} \left(\frac{3}{3x + 4} + 2 \log(3x + 4)\right)$$

---

VSAQ-3 : Find the derivative of f(x) = e^x(x² + 1)

$$f(x) = e^x (x^2 + 1)$$

$$f'(x) = e^x \frac{d}{dx}(x^2 + 1) + (x^2 + 1) \frac{d}{dx} e^x$$

$$\frac{d}{dx}(x^2 + 1) = 2x$$

$$f'(x) = e^x (2x) + (x^2 + 1) e^x$$

---

VSAQ-4 : If f(x) = e^2xlogx, (x>0) then find f'(x)

$$f(x) = e^{2x} \log x$$

The product rule states that the derivative of a product 𝑢𝑣uv is uv′+vu′, where 𝑢u and 𝑣v are functions of x.

$$f'(x) = e^{2x} \frac{d}{dx}(\log x) + \log x \frac{d}{dx}(e^{2x})$$

$$\frac{d}{dx}(\log x) = \frac{1}{x}$$

$$\frac{d}{dx}(e^{2x}) = e^{2x} \cdot 2$$

$$f'(x) = e^{2x} \left(\frac{1}{x}\right) + \log x \cdot (e^{2x} \cdot 2)$$

$$f'(x) = e^{2x} \left(\frac{1}{x} + 2 \log x\right)$$

---

VSAQ-5 : Find the derivative of sinmx.cosnx

$$f(x) = \sin(mx) \cos(nx)$$

$$\frac{d}{dx}(\sin(mx) \cos(nx)) = \sin(mx) \frac{d}{dx}(\cos(nx)) + \cos(nx) \frac{d}{dx}(\sin(mx))$$

$$\frac{d}{dx}(\cos(nx)) = -n \sin(nx)$$

$$\frac{d}{dx}(\sin(mx)) = m \cos(mx)$$

$$\frac{d}{dx}(\sin(mx) \cos(nx)) = \sin(mx)(-n \sin(nx)) + \cos(nx)(m \cos(mx))$$

$$\frac{d}{dx}(\sin(mx) \cos(nx)) = -n \sin(mx) \sin(nx) + m \cos(mx) \cos(nx)$$

---

VSAQ-6 : Find the derivative of 5sinx + e^xlogx

$$f(x) = 5 \sin x + e^x \log x$$

$$\frac{d}{dx}(5 \sin x + e^x \log x) = 5 \frac{d}{dx}(\sin x) + \frac{d}{dx}(e^x \log x)$$

$$\frac{d}{dx}(5 \sin x + e^x \log x) = 5 \cos x + [e^x \frac{1}{x} + \log x \cdot e^x]$$

$$\frac{d}{dx}(5 \sin x + e^x \log x) = 5 \cos x + e^x \left(\frac{1}{x} + \log x\right)$$

---

VSAQ-7 : Find the derivative of 5^x + logx + x³e^x

$$\frac{d}{dx} \left(5x + \log x + x^3 e^x\right) = 5 + \frac{1}{x} + 3x^2 e^x + x^3 e^x$$

---

VSAQ-8 : Find the derivative of e^x + sinxcosx

$$\frac{d}{dx} \left(e^x + \sin x \cdot \cos x\right)$$

$$= \frac{d}{dx} (e^x) + \frac{d}{dx} (\sin x \cdot \cos x)$$

$$= e^x + \sin x \frac{d}{dx}(\cos x) + \cos x \frac{d}{dx}(\sin x)$$

$$= e^x + \sin x (-\sin x) + \cos x (\cos x)$$

$$= e^x – \sin^2 x + \cos^2 x$$

$$= e^x + (\cos^2 x – \sin^2 x) = e^x + \cos 2x$$

---

VSAQ-9 : If f(x) = xe^x sin x then find f'(x)

$$\frac{d}{dx}(uvw) = uv \frac{d}{dx} w + uw \frac{d}{dx} v + vw \frac{d}{dx} u$$

$$f'(x) = \frac{d}{dx}(x e^x \sin x)$$

$$= x e^x \frac{d}{dx}(\sin x) + x \sin x \frac{d}{dx}(e^x) + e^x \sin x \frac{d}{dx}(x)$$

$$= x e^x (\cos x) + x \sin x (e^x) + e^x \sin x (1)$$

$$= x e^x \cos x + x \sin x e^x + e^x \sin x$$

$$= e^x (x \cos x + x \sin x + \sin x)$$

---

VSAQ-10 : If y = x²e^xsinx, then find dy/dx

$$\frac{d}{dx}(uvw) = uv \frac{d}{dx} w + uw \frac{d}{dx} v + vw \frac{d}{dx} u$$

$$\frac{dy}{dx} = x^2 e^x \frac{d}{dx}(\sin x) + x^2 \sin x \frac{d}{dx}(e^x) + e^x \sin x \frac{d}{dx}(x^2)$$

$$= x^2 e^x (\cos x) + x^2 \sin x (e^x) + e^x \sin x (2x)$$

$$= x^2 e^x \cos x + x^2 \sin x e^x + 2x e^x \sin x$$

$$= e^x (x^2 \cos x + x^2 \sin x + 2x \sin x)$$

---

VSAQ-11 : Find the derivative of x²n^xlog(nx), (x>0, n ∈ N)

$$\frac{d}{dx}(x^n n^{x \log x}) = x^n n^{x \log x} \frac{d}{dx}(\log(n x)) + n^{x \log x} \log(n x) \frac{d}{dx}(x^n) + x^n \log(n x) \frac{d}{dx}(n^{x \log x})$$

$$= x^n n^{x \log x} \cdot \frac{1}{x} \cdot n + n^{x \log x} \log(n x) \cdot n x^{n-1} + x^n \log(n x) \cdot n^{x \log x} \log n (\log x + 1)$$

$$= n^{x \log x} x^{n-1} (n + n x \log(n x) + x \log(n x) \log n (\log x + 1))$$

$$= n^{x \log x} x^{n-1} (n + n x \log(n x) + x \log(n x) \log n (\log x + 1))$$

---

VSAQ-12 : If y = 2x + 3/4x + 5 then find dy/dx

$$\frac{d}{dx} \left(\frac{u}{v}\right) = \frac{v \frac{d}{dx}(u) – u \frac{d}{dx}(v)}{v^2}$$

$$\frac{dy}{dx} = \frac{(4x + 5) \frac{d}{dx}(2x + 3) – (2x + 3) \frac{d}{dx}(4x + 5)}{(4x + 5)^2}$$

$$= \frac{(4x + 5)(2) – (2x + 3)(4)}{(4x + 5)^2}$$

$$= \frac{8x + 10 – 8x – 12}{(4x + 5)^2}$$

$$= \frac{-2}{(4x + 5)^2}$$

---

VSAQ-13 : If y = a – x/a + x, (x ≠ -a) then find dy/dx

$$\frac{d}{dx} \left(\frac{u}{v}\right) = \frac{v \frac{d}{dx}(u) – u \frac{d}{dx}(v)}{v^2}$$

$$\frac{dy}{dx} = \frac{(a + x) \frac{d}{dx}(a – x) – (a – x) \frac{d}{dx}(a + x)}{(a + x)^2}$$

$$= \frac{(a + x)(-1) – (a – x)(1)}{(a + x)^2}$$

$$= \frac{-a – x – a + x}{(a + x)^2}$$

$$= \frac{-2a}{(a + x)^2}$$

---

VSAQ-14 : If f(x) = sin(logx), (x > 0) then find f'(x)

$$f'(x) = \frac{d}{dx}(\sin(\log x))$$

$$= \cos(\log x) \cdot \frac{d}{dx}(\log x)$$

$$= \cos(\log x) \cdot \frac{1}{x}$$

$$= \frac{\cos(\log x)}{x}$$

---

VSAQ-15 : Find the derivative of y = log₇ (log x)

$$y = \frac{\log_e(\log x)}{\log_e(7)} = \log_7 e \cdot \log_e(\log x)$$

$$\frac{dy}{dx} = \log_7 e \cdot \frac{d}{dx}(\log_e(\log x))$$

$$= \log_7 e \cdot \frac{1}{\log x} \cdot \frac{d}{dx}(\log x)$$

$$= \log_7 e \cdot \frac{1}{\log x} \cdot \frac{1}{x}$$

$$= \frac{\log_7 e}{x \log x}$$

---

VSAQ-16 : Find the derivative of log(secx + tanx)

$$y = \log(\sec x + \tan x)$$

$$\frac{dy}{dx} = \frac{d}{dx} \log(\sec x + \tan x)$$

$$= \frac{1}{\sec x + \tan x} \cdot \frac{d}{dx}(\sec x + \tan x)$$

$$= \frac{1}{\sec x + \tan x} \cdot (\sec x \tan x + \sec^2 x)$$

$$= \frac{\sec x (\tan x + \sec x)}{\sec x + \tan x}$$

$$= \sec x$$

---

VSAQ-17 : Find the derivative of log(sin(logx))

$$y = \log(\sin(\log x))$$

$$\frac{dy}{dx} = \frac{d}{dx} \log[\sin(\log x)]$$

$$= \frac{1}{\sin(\log x)} \cdot \frac{d}{dx}(\sin(\log x))$$

$$= \frac{1}{\sin(\log x)} \cdot \cos(\log x) \cdot \frac{1}{x}$$

$$= \frac{\cos(\log x)}{\sin(\log x)} \cdot \frac{1}{x} = \frac{\cot(\log x)}{x}$$

---

VSAQ-18 : Find the derivative of y = e^Sin-1x

$$y = e^{\sin^{-1}(x)}$$

$$\frac{dy}{dx} = \frac{d}{dx} \left(e^{\sin^{-1}(x)}\right)$$

$$= e^{\sin^{-1}(x)} \cdot \frac{1}{\sqrt{1-x^2}}$$

$$= \frac{e^{\sin^{-1}(x)}}{\sqrt{1-x^2}}$$

---

$$y = e^{a \sin^{-1}(x)}$$

$$\frac{dy}{dx} = \frac{d}{dx} \left(e^{a \sin^{-1}(x)}\right)$$

$$= e^{a \sin^{-1}(x)} \cdot a \cdot \frac{1}{\sqrt{1-x^2}}$$

$$= \frac{a y}{\sqrt{1-x^2}}$$

---

VSAQ-20 : If y = Sin^-1(cos x) then find dy/dx

$$y = \sin^{-1}(\cos x)$$

$$\frac{dy}{dx} = \frac{d}{dx} \sin^{-1}(\cos x)$$

$$= \frac{1}{\sqrt{1 – (\cos x)^2}} \cdot (-\sin x)$$

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

$$= -\frac{\sin x}{|\sin x|}$$

$$= -1$$

---

VSAQ-21 : If y = Tan^-1(logx) then find dy/dx

$$y = \tan^{-1}(\log x)$$

$$\frac{dy}{dx} = \frac{d}{dx} \tan^{-1}(\log x)$$

$$= \frac{1}{1 + (\log x)^2} \cdot \frac{1}{x}$$

$$= \frac{1}{x(1 + (\log x)^2)}$$

---

VSAQ-22 : Find the derivative of Tan^-1(2x/1 – x²)

$$x = \tan \theta \quad \text{then} \quad \theta = \tan^{-1} x$$

$$\tan^{-1} \left(\frac{2x}{1 – x^2}\right) = \tan^{-1} \left(\frac{2\tan \theta}{1 – \tan^2 \theta}\right)$$

$$= \tan^{-1}(\tan 2\theta)$$

$$= 2\theta = 2(\tan^{-1} x)$$

$$\frac{d}{dx} \left(2 \tan^{-1} x\right) = 2 \frac{d}{dx} \tan^{-1} x$$

$$= 2 \cdot \frac{1}{1 + x^2}$$

---

VSAQ-23 : Find the derivative of Sin^-1(2x/1 + x²)

$$x = \tan \theta \quad \text{then} \quad \theta = \tan^{-1} x$$

$$\sin^{-1} \left(\frac{2x}{1 + x^2}\right) = \sin^{-1} \left(\frac{2\tan \theta}{1 + \tan^2 \theta}\right)$$

$$= \sin^{-1}(\sin 2\theta)$$

$$= 2\theta = 2(\tan^{-1} x)$$

$$\frac{d}{dx} \left(2 \tan^{-1} x\right) = 2 \frac{d}{dx} \tan^{-1} x$$

$$= 2 \cdot \frac{1}{1 + x^2}$$

---

VSAQ-24 : Find the derivative of Sin^-1(3x – 4x³)

$$x = \sin \theta \quad \text{then} \quad \theta = \sin^{-1} x$$

$$\sin^{-1} (3x – 4x^3) = \sin^{-1} (3\sin \theta – 4\sin^3 \theta)$$

$$= \sin^{-1}(\sin 3\theta)$$

$$= 3\theta = 3(\sin^{-1} x)$$

$$\frac{d}{dx} \left(3 \sin^{-1} x\right) = 3 \frac{d}{dx} \sin^{-1} x$$

$$= 3 \cdot \frac{1}{\sqrt{1 – x^2}}$$

---

VSAQ-25 : Find the derivative of Cos^-1(4x³ – 3x)

$$x = \cos \theta \quad \text{then} \quad \theta = \cos^{-1} x$$

$$\cos^{-1} (4x^3 – 3x) = \cos^{-1} (4\cos^3 \theta – 3\cos \theta)$$

$$= \cos^{-1}(\cos 3\theta)$$

$$= 3\theta = 3(\cos^{-1} x)$$

$$\frac{d}{dx} \left(3 \cos^{-1} x\right) = 3 \frac{d}{dx} \cos^{-1} x$$

$$= 3 \cdot \left(-\frac{1}{\sqrt{1 – x^2}}\right)$$

$$= -\frac{3}{\sqrt{1 – x^2}}$$

---

VSAQ-26 : If y = Sin^-1 √x, find dy/dx

$$y = \sin^{-1}(\sqrt{x})$$

$$\frac{dy}{dx} = \frac{d}{dx}(\sin^{-1}(\sqrt{x}))$$

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

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

---

VSAQ-27 : Find the derivative of Sinh^-1(3x/4)

$$y = \sinh^{-1}\left(\frac{3x}{4}\right)$$

$$\frac{dy}{dx} = \frac{d}{dx}\left(\sinh^{-1}\left(\frac{3x}{4}\right)\right)$$

$$= \frac{1}{\sqrt{1 + \frac{9x^2}{16}}} \cdot \frac{3}{4}$$

$$= \frac{3}{4} \cdot \frac{1}{\sqrt{\frac{16 + 9x^2}{16}}}$$

$$= \frac{3}{4} \cdot \frac{1}{\sqrt{\frac{16 + 9x^2}{16}}} = \frac{3}{4} \cdot \frac{4}{\sqrt{16 + 9x^2}} = \frac{3}{\sqrt{16 + 9x^2}}$$

---

VSAQ-28 : Find the derivative of y = √2x – 3 + √7 – 3x

$$y = \sqrt{2x – 3} + \sqrt{7 – 3x}$$

$$\frac{dy}{dx} = \frac{d}{dx}\left(\sqrt{2x – 3} + \sqrt{7 – 3x}\right)$$

$$\frac{d}{dx}(\sqrt{2x – 3}) = \frac{1}{2\sqrt{2x – 3}} \cdot 2$$

$$\frac{d}{dx}(\sqrt{7 – 3x}) = \frac{1}{2\sqrt{7 – 3x}} \cdot (-3)$$

$$= \frac{1}{\sqrt{2x – 3}}$$

$$= -\frac{3}{2\sqrt{7 – 3x}}$$

$$\frac{dy}{dx} = \frac{1}{\sqrt{2x – 3}} – \frac{3}{2\sqrt{7 – 3x}}$$

---

VSAQ-29 : Find the derivative of 7x³ + 3x

$$\frac{d}{dx} (7x^3 + 3x) = 7x^3 + 3x (\log 7) \cdot \frac{d}{dx}(x^3 + 3x)$$

$$\frac{d}{dx}(x^3 + 3x) = 3x^2 + 3$$

$$= 7x^3 + 3x (\log 7)(3x^2 + 3)$$

---

VSAQ-30 : Find the derivative of (x³ + 6x² + 12x – 13)¹⁰⁰

$$y = (x^3 + 6x^2 + 12x – 13)^{100}$$

$$\frac{d}{dx} (x^3 + 6x^2 + 12x – 13)^{100}$$

$$= 100(x^3 + 6x^2 + 12x – 13)^{99} \cdot \frac{d}{dx}(x^3 + 6x^2 + 12x – 13)$$

$$\frac{d}{dx}(x^3 + 6x^2 + 12x – 13) = 3x^2 + 12x + 12$$

$$= 100(x^3 + 6x^2 + 12x – 13)^{99} \cdot (3x^2 + 12x + 12)$$

---

VSAQ-31 : If y = (Cot^-1x³)² then find dy/dx

$$y = (\cot^{-1}(x^3))^2$$

$$\frac{dy}{dx} = \frac{d}{dx}((\cot^{-1}(x^3))^2)$$

$$= 2(\cot^{-1}(x^3)) \cdot \frac{d}{dx}(\cot^{-1}(x^3))$$

$$\frac{d}{dx}(\cot^{-1}(x^3)) = -\frac{1}{1 + (x^3)^2} \cdot 3x^2$$

$$= 2(\cot^{-1}(x^3)) \cdot \left(-\frac{3x^2}{1 + x^6}\right)$$

$$= -6x^2 (\cot^{-1}(x^3)) \frac{1}{1 + x^6}$$

---

VSAQ-32 : Find dy/dx if 2x² – 3xy + y² + x + 2y – 8 = 0

$$2x^2 – 3xy + y^2 + x + 2y – 8 = 0$$

$$4x – 3(x \frac{dy}{dx} + y) + 2y \frac{dy}{dx} + 1 + 2 \frac{dy}{dx} = 0$$

$$-3x \frac{dy}{dx} + 2y \frac{dy}{dx} + 2 \frac{dy}{dx} = -4x + 3y – 1$$

$$\frac{dy}{dx} (2y – 3x + 2) = 3y – 4x – 1$$

$$\frac{dy}{dx} = \frac{3y – 4x – 1}{2y – 3x + 2}$$

---

VSAQ-33 : Find dy/dx if x = acos³t, y = asin³t

$$\frac{dx}{dt} = \frac{d}{dt}(a \cos(3t)) = a \cdot \frac{d}{dt}(\cos(3t))$$

$$= a \cdot (-3 \sin(3t)) = -3a \sin(3t)$$

$$\frac{dy}{dt} = \frac{d}{dt}(a \sin(3t)) = a \cdot \frac{d}{dt}(\sin(3t))$$

$$= a \cdot (3 \cos(3t)) = 3a \cos(3t)$$

$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3a \cos(3t)}{-3a \sin(3t)}$$

$$= \frac{\cos(3t)}{-\sin(3t)} = -\cot(3t)$$

---

VSAQ-34 : Find the derivative of x^x

$$y = x^x$$

$$\log y = \log (x^x)$$

$$\log y = x \log x$$

$$\frac{1}{y} \frac{dy}{dx} = x \cdot \frac{1}{x} + \log x \cdot 1$$

$$\frac{1}{y} \frac{dy}{dx} = 1 + \log x$$

$$\frac{dy}{dx} = y(1 + \log x)$$

$$\frac{dy}{dx} = x^x (1 + \log x)$$

---

VSAQ-35 : If y = x⁴ + tanx then find y”

$$y = x^4 + \tan x$$

$$y’ = 4x^3 + \sec^2 x$$

$$y” = 12x^2 + 2 \sec^2 x \tan x$$

---

VSAQ-36 : If y = ae^nx + be^-nx then prove that y” = n²y

$$y = ae^{nx} + be^{-nx}$$

$$y’ = ane^{nx} + b(-n)e^{-nx}$$

$$y’ = ane^{nx} – bne^{-nx}$$

$$y” = an^2e^{nx} – bn^2e^{-nx}$$

$$y” = n^2(ae^{nx} + be^{-nx}) = n^2y$$

---

VSAQ-37 : If f(x) = 2x² + 3x – 5 then prove that f'(0) + 3f'(-1) = 0

$$f(x) = 2x^2 + 3x – 5$$

$$f'(x) = 4x + 3$$

$$f'(0) = 4 \cdot 0 + 3 = 3$$

$$f'(-1) = 4 \cdot (-1) + 3 = -4 + 3 = -1$$

$$f'(0) + 3f'(-1) = 3 + 3 \cdot (-1) = 3 – 3 = 0$$

---

VSAQ-38 : If f(x) = 1 + x + x² + … + x¹⁰⁰, then find f'(1)

$$f(x) = 1 + x + x^2 + \ldots + x^{100}$$

$$f'(x) = 0 + 1 + 2x + 3x^2 + \ldots + 100x^{99}$$

$$f'(1) = 1 + 2 + 3 + \ldots + 100$$

$$f'(1) = \frac{100 \times 101}{2} = 5050$$

---

VSAQ-39 : If x³ + y³ – 3xy = 0 then find dy/dx

$$x^3 + y^3 – 3xy = 0$$

$$3x^2 + 3y^2 \frac{dy}{dx} – 3(x \frac{dy}{dx} + y) = 0$$

$$3x^2 – 3y + (3y^2 – 3x)\frac{dy}{dx} = 0$$

$$(y^2 – x)\frac{dy}{dx} = y – x^2$$

$$\frac{dy}{dx} = \frac{y – x^2}{y^2 – x}$$

---

VSAQ-40 : If x = logt + sint, y = e^t + cost then find dy/dx

$$\frac{dy}{dx} = \frac{e^t – \sin t}{\frac{1}{t} + \cos t}$$

$$\frac{dy}{dx} = \frac{t(e^t – \sin t)}{1 + t \cos t}$$

---

VSAQ-41 : Find the derivative of x = a(cos t + t sin t), y = a(sin t – t cos t)

$$x = a(\cos t + t \sin t)$$

$$\frac{dx}{dt} = a(-\sin t + t \cos t + \sin t) = a t \cos t$$

$$y = a(\sin t – t \cos t)$$

$$\frac{dy}{dt} = a(\cos t – t \sin t) = at \sin t$$

$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{at \sin t}{at \cos t} = \tan t$$

---

VSAQ-42 : If x = 3cost – 2cos3t, y = 3sint – 2sin3 t then find dy/dx

$$x = 3 \cos t – 2 \cos 3t$$

$$\frac{dx}{dt} = -3 \sin t + 6 \cos^2 t \sin t = 3 \sin t (2 \cos^2 t – 1) = 3 \sin t \cos 2t$$

$$y = 3 \sin t – 2 \sin 3t$$

$$\frac{dy}{dt} = 3 \cos t – 6 \sin^2 t \cos t = 3 \cos t (1 – 2 \sin^2 t) = 3 \cos t \cos 2t$$

$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3 \cos t \cos 2t}{3 \sin t \cos 2t} = \cot t$$

---

VSAQ-43 : Differentiate f(x) = e^x w.r.to g(x) = √x

$$f(x) = e^x$$

$$f'(x) = e^x$$

$$g(x) = \sqrt{x}$$

$$g'(x) = \frac{1}{2\sqrt{x}}$$

$$\frac{df}{dg} = \frac{f'(x)}{g'(x)} = \frac{e^x}{\frac{1}{2\sqrt{x}}} = 2\sqrt{x} e^x$$