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Let y = x^{(sinx – cosx)}; log y = (sin x – cos x) 

log x diff w.r.t x. 

\(\frac{1}{y}\frac{dy}{dx}\) = (sinx – cosx) .\(\frac{1}{x}\) + logx (cosx + sinx); 

\(\frac{dy}{dx}\) = y [\(\frac{sinx - cosx}{x}\) + log x.(cosx + sinx)]

Given x = a cos⁴θ, y = a sin 4θ 

\(\frac{dx}{d\theta}\) = a(4cos³θ( – sinθ)) \(\frac{dy}{d\theta}\) = 4a sin³θcosθ

∴ \(\frac{dy}{dx}\) = \(\frac{4asin^3\theta cos\theta}{-4acos^3\theta sin\theta}\) = - tan²θ

Given y = sinmx 

\(\frac{dy}{dx}\) = -mcosmx 

\(\frac{d^2y}{dx^2}\) = m²sinmx = -m²y

∴ \(\frac{d^2y}{dx^2}\) + m²y = 0

Let y = (x + cos x)(x – tanx) 

\(\frac{dy}{dx}\)= (x + cosx)(1 – sec²x) + (xtanx)(1 – sinx).

Let y = (sin x)x , Taking logm 

logy = xlogsinx

\(\frac{1}{y}\frac{dy}{dx}\)= x . \(\frac{1}{sinx}\). cosx + logsinx.1 

\(\frac{dy}{dx}\)= y[xcotx + logsinx]

Let y = x^sinx, taking log^m both sides 

log y = sinx log x, Diff w.r.t x 

\(\frac{1}{y}.\frac{dy}{dx}\)= sin x .\(\frac{1}{x}\) + log x . cos x 

\(\frac{dy}{dx}\) = y[\(\frac{sinx}{x}\) + log x . cos x]

Let u = sin²x, v = x² 

Differentiate both w.r.t. x 

\(\frac{du}{dx}\) = 2 sin x. cos x, 

\(\frac{dv}{dx}\) = 2x

∴ \(\frac{du}{dv}\) = \(\frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{2sinxcosx}{2x}\)

= \(\frac{sinxcosx}{x}\)

Let y = 7^sin√x

\(\frac{dy}{dx}\) = 7^sin√x.log 7^. cos√x . \(\frac{1}{2\sqrt x}\)

Let y = sin x . sin2x 

\(\frac{dy}{dx}\)= sinx(2cos2x) + sin2xcosx = 2sinxcos2x + cosx sin2x

Given y = 4^x+y, diff. w r.t. x

\(\frac{dy}{dx}\) = 4^{x + y}log4(1 + \(\frac{dy}{dx}\)) = 4^{x + y}

\(\frac{dy}{dx}\)(1 - 4^{x + y}log4) = 4^{x + y} log4

∴ \(\frac{dy}{dx}\) = \(\frac{4^{x + y}.log4}{1 - 4^{x + y}.log4}\)

y = e^ax + e^-ax 

y₁ = ae^ax – ae^-ax 

y₂ = a²e^ax + a²e^-ax 

= a² (e^ax + e^-ax) = a²y 

∴ y² – a²y = 0.

Let y = log_e e (constant)

\(\frac{dy}{dx} = 0\)

Let y = 5e^x – log x – 3√x

\(\frac{dy}{dx} = 5e^x - \frac{1}{x} - 3\frac{1}{2\sqrt x}\)

According to question

x³ + y³ = 3axy

Thus at point (\(\frac{3a}{2}\), \(\frac{3a}{2}\))

Putting the coordinates in the equation

\(\frac{27a^3}{8}\) + \(\frac{27a^3}{8}\) = \(\frac{54a^3}{4}\)

\(\frac{27a^3 + 27a^3}{8}\) = \(\frac{54a^3}{4}\)

\(\frac{54a^3}{8}\) = \(\frac{54a^3}{4}\)

Thus

a = \(\frac{1}{2}\)

Thus the coordinates are (\(\frac{1}{2}\), \(\frac{1}{2}\)).

Let u = tan²x, v = cos²x 

Differentiate both w.r.t x 

\(\frac{du}{dx}\) = 2tan x. sec²x,

\(\frac{dv}{dx}\) = 2cosx(-sinx)

∴ \(\frac{du}{dv}\) = \(\frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{2tanx.sec^2x}{2sinx.cosx}\)

= \(\frac{1}{sec^4x}\)

Let y = \(\frac{4x^2}{x} - \frac{3x}{x}\) = 4x – 3 

∴ \(\frac{dy}{dx}\) = 4(1) – 0 = 4

Let y = \(\Big(\sqrt x + \frac{1}{\sqrt x}\Big)^2\)

= x + \(\frac{1}{x}\) + 2

\(\frac{dy}{dx}\) = 1  - \(\frac{1}{x^2}\) + 0 = 1 - \(\frac{1}{x^2}\)

Given y = x + tan x

\(\frac{dy}{dx}\) = 1 + sec²x = 1 + \(\frac{1}{cos^2x}\)

= \(\frac{cos^2x + 1}{cos^2x}\)

∴ cos²x\(\frac{dy}{dx}\) = cos²x + 1 = 1 – sin²x + 1 

= 2 – sin²x 

∴ cos²x \(\frac{dy}{dx}\) = 2 – sin²x

Let y = (sin x)² 

\(\frac{dy}{dx}\)= 2 sin x .(cosx) = sin2x

Let y(a² – x²)¹⁰

\(\frac{dy}{dx}\) = 10(a² – x²)^{10 – 1} . \(\frac{dy}{dx}\)(a² – x²)

= 10(a² – x²)⁹(-2x) = -20x(a² – x²)⁹

Let y=(x – a) (x – b) 

\(\frac{dy}{dx}\) = (x – a) . 1 + (x – b) . 1 

= x – a + x – b 

= 2x – a – b

Let y = x⁵ (3 – 6x^-9)

\(\frac{dy}{dx}\) = x⁵(+54x^-10) + (3 – 6x^-9)(5x⁴) 

= 54x^-5 + 15x⁴ – 30x^-5 

= 24x^-5 + 15x⁴

Let y = log x (log log(x))

\(\frac{dy}{dx}\) = \(\frac{1}{log(logx)}.\frac{1}{logx}.\frac{1}{x}\)

Let y = sin³(√x) = (sin√x))³ 

\(\frac{dy}{dx}\) = 3.sin² √x. cos√x .\(\frac{1}{2\sqrt x}\).

Let y = x^-3 (5 + 3x) 

\(\frac{dy}{dx}\)= x^-3 (3) + (5 + 3x) (-3 . x^-3-1 ) 

= 3x^-3 – 15x^-4 - 9x^-3 

= -15x^-4 – 6x^-3

⁼ \(\frac{-3}{x^4}(2x + 5)\)

Let y = log(log (tan x))

\(\frac{dy}{dx}\) = \(\frac{1}{log(tanx)}.\frac{1}{tanx} . sec^2x\)

Let y = (5x² + 3x – 1) ( x – 1) 

\(\frac{dy}{dx}\)= (5x² + 3x – 1) \(\frac{d}{dx}\)(x – 1) + (x – 1) \(\frac{d}{dx}\)(5x² + 3x – 1) 

= (5x² + 3x – 1) + (x – 1) (10x + 3) 

= 5x² + 3x – 1 + 10x² – 10x + 3x – 3 

= 15x² – 4x – 4

Let y = [log(cos x)]² 

\(\frac{dy}{dx}\) = 2log(cos x). \(\frac{1}{cosx}\)(-sin x) 

= -2 tan x log(cos x)

Let y = cosx³ 

\(\frac{dy}{dx}\)= -sinx³ .3x²

Let y = 3x² .logx.

\(\frac{dy}{dx}\) = 3x² .\(\frac{1}{x}\) + logx . 3x² . log_e3 . 2x

Given sin xy = cos(x + y), diff w.r.t x. 

cos(xy) \(\Big[x\frac{dy}{dx} + y\Big]\) 

\(\frac{dy}{dx}\)[sin(x + y) + xcos (xy)] = -sin(x + y) – ycos(xy)

\(\frac{dy}{dx} = \frac{-\Big[sin(x + y) + cosxy\Big]}{\Big(sin(x + y) + xcosxy\Big)}\)

Let y = cos3x . sin5x (Trans using formula) 

sy = 2[sin8x – sin(-2x)] = 2 [sin8x] + 2sin2x 

\(\frac{dy}{dx}\) = 16cos8x + 4cos2x 

OR Let y = cos 3x . sin 5x 

\(\frac{dy}{dx}\) = cos3x(5cos5x) + sin5x(-3sin3x) = 5cos3x sin5x – 3sin5x.sin3x.

Let y = cos⁵x . cos(x⁵) 

\(\frac{dy}{dx}\) = cos⁵x(-sin(x⁵)5x⁴) + cos(x⁵).5cos⁴x.(-sinx) 

= – cos⁵x sin(x⁵) 5x⁴ – 5 cos(x⁵) . cos⁴x . sinx

Let y = e^2x.sin3x 

\(\frac{dy}{dx}\) = e^2x(3cos3x) + sin3x(2e^2x)

For the following problems chain rule to be used: 

\(\frac{d}{dx}\)f(g(x)) = f'(g(x)).g'(x) 

\(\frac{d}{dx}\)[f(x)]ⁿ = n[f(x)]^{n - 1} x \(\frac{d}{dx}\)f(x)

(i) Let y = sin² x = (sin x)² 

\(\frac{dy}{dx}\) = 2(sin x)^2-1 \(\frac{d}{dx}\)(sin x) 

= 2 sin x (cos x) 

= sin 2x

(ii) y = cos² x = (cos x)² 

\(\frac{dy}{dx}\) = 2(cos x)^2-1 \(\frac{d}{dx}\)(cos x) 

= 2 cos x (-sin x) 

= -2 sin x cos x 

= -sin 2x

(iii) y = cos³ x 

y = (cos x)³

\(\frac{dy}{dx}\) = 3(cos x)^{3 - 1} \(\frac{d}{dx}\)(cos x) 

= 3 cos² x (-sin x) 

= -3 cos² x sin x 

= -3 cos x (sin x cos x) [Multiply and divide by 2] 

= \(\frac{-3}{2}\) cos x (2 sin x cos x) 

= \(\frac{-3}{2}\) cos x sin 2x

(a) f(x) = a^x, a > 1

(a) 1 (By formula)

(c) (0, 1)

y = e^x 

Put x = 0, we get y = e⁰ = 1. 

∴ The graph intersects the y-axis at (0, 1)

f(x) = x³ - \(\frac{1}{x^3}\) ... (1)

f(\(\frac{1}{x}\)) = (\(\frac{1}{x}\))³ - \(\frac{1}{(\frac{1}{x})^3}\) = \(\frac{1}{x^3}\) - x³ ... (2)

(1) + (2) gives f(x) + f(\(\frac{1}{x}\))

= x³ - \(\frac{1}{x^3}+\frac{1}{x^3}-x^3\) = 0

Hence Proved.

(a) 2

\(f(x) = \begin{cases} x^2-4x & \quad \text{if } x≥0 \text{}\\ x+2 & \quad \text{if } x<2 \text{} \end{cases}\)

f(0) = 0 + 2 = 2 

[For x = 0 take f(x) = x + 2]

(c) 1 (By formula)

(a) \(-\frac{1}{x^2}\)

For a polynomial function to be an odd function each term should have odd powers pf x. 

Therefore there should not be an even power of x term.

∴ k = 0.

(b) 1

When x = 1, \(\frac{x+2}{x-1}\) is not defined.

(c) 5

\(f(x) = \begin{cases} x^2-4x & \quad \text{if } x≥2 \text{}\\ x+2 & \quad \text{if } x<2 \text{} \end{cases}\)

f(5) = 5² – 4(5) = 25 – 20 = 5 

[For x = 5 take f(x) = x² – 4x]

(d) x² + x + 1

f(x) = x² – x + 1 

f(x + 1) = (x + 1)² – (x + 1) + 1

= x² + 2x + 1 – x – 1 + 1 

= x² + x + 1

(b) [0, ∞)

[0, ∞) since in this interval 0 is included and f(0) = 0.

(d) constant function

y = sin(log x) 

y₁ = cos(log x) \(\frac{d}{dx}\)(log x) 

y₁ = cos(log x).\(\frac{1}{x}\) 

∴ xy₁ = cos(log x)

Differentiating both sides with respect to x, we get 

xy₂ + y₁(1) = -sin(log x).\(\frac{1}{x}\) 

⇒ x[xy₂ + y₁] = -sin(log x) 

⇒ x² y₂ + xy₁ = -y 

⇒ x² y₂ + xy₁ + y = 0

\(\lim\limits_{x \to a}\frac{x^9-a^9}{x-a}\) = \(\lim\limits_{x \to 3}(x+6)\)

9.a^{9 - 1} = 3 + 6 

9.a⁸ = 9 

a⁸ = 1 

Taking squareroot on bothsides, we get 

\((a^8)^{\frac{1}{2}}\)= ±1 

a⁴ = ±1 

But a⁴ = -1 is imposssible. 

∴ a⁴ = 1 

Again taking squareroot, we get 

\((a^4)^{\frac{1}{2}}\)= ±1 

a² = ±1 

a² = -1 is imposssible 

∴ a² = 1 

Again taking positive squareroot, a = ±1