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As we know that area of a circular disc = πr² and circumference of a circular disc = 2πr

Here r = radius of circular disc

To find dA/dC here A = area of circular disc and c = circumference of the circular disk.

dA/dC = (dA/dr) x (dr/dC) 

dA/dr = 2πr

dC/dr = 2π

dA/dC = 2πr/2π = r

(dA/dr)_{at x = 3} = 3cm

Let,

y = (sec x − 1) (sec x + 1) 

= sec²x − 1 

= tan²x 

∴ \(\frac{dy}{dy}=\frac{d}{dx}\,(tan^2\,x)\) 

\(=2\,tan\,x\frac{d}{dx}\,tan\,x\) 

\(=2\,tan\,x.sec^2\,x.\)

Given, 

f(x) = (x² + 6x − 5)(1 − x)

= x² + 6x − 5−x³ − 6x² + 5x

= -x³ + 5x² +11x - 5

\(∴ f'(x)=\frac{d}{dx}(-x^3-5x^2+11x+5)\) 

= −3x² − 5.2x + 11.1 − 0

= −3x² − 10x + 11

∴ At x = 3,

f'(x) = 3.3² − 10.3 + 11

= −27 − 30 + 11

= −57 + 11

= −46

∴ Slope of the tangent to f(x) at x = 3 is – 46.

 Let,

\(f(x)=\sqrt\frac{1-cos\,2x}{1+cos\,2x}\) 

\(=\sqrt\frac{(1-cos\,2x)(1+cos\,2x)}{(1+cos\,2x)^2}\) 

\(=\sqrt\frac{(1-cos^2\,2x)}{(1+cos\,2x)^2}\) 

\(=\sqrt\frac{sin^2\,2x}{(1+cos\,2x)^2}\) 

\(=\sqrt\frac{sin^2\,2x}{(1+cos\,2x)^2}\) 

\(=\frac{sin\,2x}{1+cos\,2x}\)  

\(∴ f'(x)=\frac{d}{dx}\{\frac{sin2x}{1+cos2x}\}\) 

\(=\frac{(1+cos2x)\frac{d}{dx}(sin2x)-sin2x\frac{d}{dx}(1+cos2x)}{(1+cos2x)^2}\) 

\(=\frac{(1+cos2x)(2cos2x)-sin2x(-2sin2x)}{(1+cos2x)^2}\) 

\(=\frac{2cos2x+2cos^22x+2sin^22x}{(1+cos2x)^2}\) 

\(=\frac{2(1+cos2x)}{(1+cos2x)^2}\) 

\(=\frac{2}{1+cos2x}\) 

\(=\frac{2}{2\,cos^2x}\) 

\(=sec^2x.\) 

Alternative method.

\(f(x)=\sqrt\frac{1-cos\,2x}{1+cos\,2x}\) 

\(f(x)=\sqrt\frac{2sin^2x}{2cos^2x}\) 

\(f(x)=tan\,x\) 

\(f'(x)=\frac{d}{dx}(tan\,x)\) 

\(=\sec^2x.\) 

Given,

\(y=\sqrt x+\frac{1}{\sqrt{x}}\)

⇒ \(\frac{dy}{dx}=\frac{d}{dx}(\sqrt x+\frac{1}{\sqrt x})\) 

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

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

∴ L.H.S 

= \(2x\frac{dy}{dx}+y\) 

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

\(=\sqrt{x}-\frac{1}{\sqrt{x}}+\sqrt{x}+\frac{1}{\sqrt{x}}\) 

\(=2\sqrt{x}\) 

= R.H.S

Let f(x) = sin 2x

∴ By first principle

\(f'(x)= \lim\limits_{h \to 0}\frac{f(x+h)-f(x)}{h}\)

\(= \lim\limits_{h \to 0}\frac{sin2(x+h)-sin2x}{h}\)

\(= \lim\limits_{h \to 0}\frac{2cos\frac{2(x+h)+2x}{2}sin\frac{2(x+h)-2x}{2}}{h}\)

= \( \lim\limits_{h \to 0}\frac{2cos(h+2x)sinh}{h}\)

= 2\(​\lim\limits_{h \to0 }cos(h+2x)\lim\limits_{h \to 0}(\frac{sinh}{h})\)

= 2∙ cos 2x ∙1

= 2 cos 2x..

Given, 

f(x) = x cos x

\(f(x)=\frac{d}{dx}(xcosx)\)

\(=cosx\frac{d}{dx}x +x\frac{d}{dx}cosx\)

= cos x∙ 1 + x ∙(− sin x)

= cos x − x sin x

f(x) = e^x sin x +xⁿ cos x

∴ f'(x) = \(\frac{d}{dx}\){e^x sin x + xⁿ cos x}

= \(\frac{d}{dx}\)(e^xsin x)+\(\frac{d}{dx}\)(xⁿcos x)

= sinx\(\frac{d}{dx}\)e^x + e^x\(\frac{d}{dx}\)sinx + cosx\(\frac{d}{dx}\)xⁿ + xⁿ\(\frac{d}{dx}\)cosx

= sin.e^x + e^x . cos x + cos x ∙ nxⁿ⁻¹ + xⁿ . (− sin x)

= e^x (sin x + cos x) + xⁿ⁻¹ [ncos x − xsin x]

Let, 

f(x) = 3^x + x³ + 4x − 5

∴ \(\frac{d}{dx}f(x)=\frac{d}{dx}f(x)\)

⟹ \(f'(x)=\frac{d}{dx}(3^x+x^3+4x-5)\)

= 3^x loge 3 + 3x² + 4 ∙1 − 0

= 3^x loge 3 + 3x² + 4.

Let,

\(f(x)=e\sqrt{tanx}\)

\(=\lim\limits_{h \to 0}\frac{f(x+h)-f(x)}{h}\) 

\(=\lim\limits_{h \to 0}\frac{e^\sqrt{tan(x+h)}-e^\sqrt{tanx}}{h}\) 

\(=\lim\limits_{h \to 0}e^\sqrt{tan(x+h)}\{\frac{e^\sqrt{tan(x+h)-tanx}-1}{h}\}\)\(e^\sqrt{tan(x+h)}\lim\limits_{h \to 0}\frac{e^\sqrt{tan(x+h)-tanx}-1}{\sqrt{tan(x+h)}-\sqrt{tanx}}\).\(\lim\limits_{h \to 0}\frac{e^\sqrt{tan(x+h)}-\sqrt{tanx}}{h}\)  

\(=e^\sqrt{tanx}.1.\lim\limits_{h \to 0}\frac{tan(x+h)-tanx}{h}\times\) \(\frac{1}{\sqrt{tan(x+h)}+\sqrt{tanx}}\)  

\(=e^\sqrt{tanx}.1.\lim\limits_{h \to 0}\frac{sin\,h}{h\,cos\,x(x+h)cos\,x}\times\)\(\frac{1}{\sqrt{tan(x+h)}+\sqrt{tanx}}\)

\(=\frac{e^\sqrt{tanx}}{2\sqrt{tan\,x}}.\frac{1}{cos^2x}\) 

\(=\frac{sec^2\,x}{2}.\frac{e^{tan\,x}}{\sqrt{tan\,x}}\)

Let, 

f(x) = cos(2x² − 3)

\(∴ f'(x)= \lim\limits_{h \to 0}\frac{f(x+h)-f(x)}{h}\) 

\(= \lim\limits_{h \to 0}\frac{cos\{2(x+h)^2-3\}-cos\{2x^2-3\}}{h}\) 

\(= \lim\limits_{h \to 0}\frac{-2sin(\frac{2(x+h)^2-3+2x^2-3}{2})\,sin(\frac{2(x+h)^2-3-2x^2+3}{2})}{h}\) 

\(=-2 \lim\limits_{h \to 0}\frac{-2sin\frac{2x^2+4xh+2x^2-6}{2}\,sin\frac{2x^2+4xh+2h^2-2x^2}{2}}{h}\) 

\(= -2\lim\limits_{h \to 0}\frac{sin(2x^2+h^2+2xh-3)\,sin(h^2+2xh)}{h}\) 

\(= -2\lim\limits_{h \to 0}(h+2x)\frac{sin(h^2+2xh)}{h^2+2xh}\,\lim\limits_{h \to 0}sin(2x^2+h^2+2xh-3)\) 

\(= -2\lim\limits_{h \to 0}(h+2x)\lim\limits_{h \to 0}\frac{sin\,(h^2+2x+h)}{h^2+2xh}sin(2x^2-3)\)

\(= -2.2x.1.sin(2x^2-3)\) 

\(= -4x\,sin(2x^2-3)\)

Let 

f(x) = log x

∴ \(f'(x)=\lim\limits_{h \to 0}\frac{f(x+h)-f(x)}{h}\)

\(=\lim\limits_{h \to 0}\frac{log(x+h)-logx}{h}\)

\(=\lim\limits_{h \to 0}\frac{log(\frac{x+h}{x})}{h}\)

\(=\lim\limits_{h \to 0}\frac{log(1+\frac{h}{x})}{h}\)

\(=\lim\limits_{h \to 0}\frac{log(1+\frac{h}{x})}{\frac{h}{x}}\times\frac{1}{x}\)

\(=1\times\frac{1}{x}\) 

= \(\frac{1}{x}\)

Let 

f(x) = x cos x

∴ \(f'(x)=\lim\limits_{h \to 0}\frac{f(x+h)-f(x)}{h}\) 

\(=\lim\limits_{h \to 0}\frac{(x+h)cos(x+h)-xcosx}{h}\)

\(=\lim\limits_{h \to 0}\frac{(x+h)[cos\,x\,cosh\,h-sinx.sinh]-xcosx}{h}\)

\(=\lim\limits_{h \to 0}\frac{xcosh.cosx-x\,sinx.sinh+h\,cosx.cosh-h\,sinx.sinh-x\,cosx}{h}\)

\(=\lim\limits_{h \to 0}\frac{xcos\,x.(cos\,h-1)+h[cos\,x.cos\,h-sin\,x.sin\,h-x\,sin\,x.sin\,h}{h}\)

\(=\lim\limits_{h \to 0}\frac{xcos\,x(cos\,h-1)}{h}+\)\(\lim\limits_{h \to 0}(-xsin\,x)(\frac{sin\,h}{h})+\) \(\lim\limits_{h \to 0}cos(x+h)\)

\(=​​-x\,cos\,x\lim\limits_{h \to 0}\frac{1-cos\,h}{h}-xsinx\lim\limits_{h \to 0}(\frac{sin\,h}{h})\) +\(+\lim\limits_{h \to 0}cos(x+h)\)   

\(= -x\,cos\,x.\lim\limits_{h \to 0}\frac{2sin^2\frac{h}{2}}{h\times\frac{h}{4}}\times\frac{h}{4}\)\(-xsinx.1+cosx\) 

\(=-x\,cos\,x.\frac{2}{4}.\lim\limits_{h \to 0}\) \((\frac{sin\frac{h}{2}}{\frac{h}{2}})^2\) \(h-x\,sin\,x+cos\,x\)

\(=-x\,cos\,x.\frac{1}{2}.1.0-xsinx+cosx\) 

= \(cos\,x - x\,sin\,x\)

Correct Answer is (A) \(\frac{3}{4}\) 

\(x=t^2;y=t^3\)

\(\frac{dy}{dt}=3t^2;\frac{dx}{dt}=2t\)

\(\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\)\(=\frac{3t}{2}\)

\(\frac{d^2y}{dx^2}=\frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}\)\(=\frac{\frac{3t}{2}}{2t}\)

\(=\frac{3}{4}\)

Given,

f(x) = ax² + bx + c

∴ f'(x) = \(\lim\limits_{h \to 0}\frac{f(x+h)-f(x)}{h}\)

= \(\lim\limits_{h \to 0}\frac{[{a(x+h)^2+b(x+h)+c}]\, -\,[{ax^2+bx+c}]}{h}\)

= \(\lim\limits_{h \to 0}\frac{a(x^2+2xh+h^2)+bx+bh-ax^2-bx}{h}\)

= \(\lim\limits_{h \to 0}\frac{2axh+ah^2+bh}{h}\)

= 2ax + a∙0 + b

= 2ax + b

Let f(x) = \(\frac{x}{sinx}\)

∴ f(x) = \(\frac{d}{dx}(\frac{x}{sinx})\)

= \(\frac{sinx\frac{d}{dx}x-x\frac{d}{dx}sinx}{(sinx)^2}\)

= \(\frac{sinx.1-x.cosx}{sin^2x}\)

= \(cosec\,x-x\,cot\,x\,cosec\,x\)

= \((1-x\,cot\,x).cosec\,x\)

Given, 

f(x) = (x−2)² (2x−3)

= (x²−4x+4)(2x−3)

= 2x³ − 8x² + 8x − 3x² + 12x − 12

= 2x³ − 11x² + 20x − 12.

f'(x) = \(\frac{d}{dx}\){2x³−11x²+20x−12}

\(=2\frac{d}{dx}x^3-11\frac{d}{dx}x^2+20\frac{d}{dx}x-\frac{d}{dx}12\) 

\(=2.3x^2-11.2x+20.1-0\)

\(=6x^2-22x+20\)

Let y = 2x⁴ + x

∴ \(\frac{dy}{dx}=\frac{d}{dx}(2x^4+x)\) 

\(=\frac{d}{dx}2x^4+\frac{d}{dx}x\)

\(=2\frac{d}{dx}2x^4+1\) 

= 2∙4∙x³ + 1

= 8x³ + 1.

Given,

\(y=1+x+x^2+...+x^{50}.\)

∴ \(\frac{dy}{dx}=\frac{d}{dx}[1+x+x^2+...+x^{50}]\) 

\(=\frac{d}{dx}1+\frac{d}{dx}x+\frac{d}{dx}x^2+...+\frac{d}{dx}x^{50}\) 

⇒ \(\frac{dy}{dx}\)= 1+2x+3x²+....+50x⁴⁹ 

∴ At  x = 1,

\(\frac{dy}{dx}\)= 1+2+3+....+50

= \(\frac{50(50+1)}{2}\)

= 25 × 51 

= 1275.

Correct Answer is (B) - n² x

Given: 

x=a cos nt-b sin nt

\(\frac{dx}{dt}\)\(\)= -an sin nt - bn cos nt

\(\frac{d^2x}{dt^2}\)= -an² cos nt + bn² sin nt

= -n² (a cos nt-b sin nt )

= - n² x

Given:

\(y=x+e^x\)

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

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

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

\(\frac{d^2x}{d^2y}=\frac{1}{e^x}\)

=e^-x

Given:

y = |x - x²|

y = \(\begin{cases}x-x^2;&x\geq0\\x^2-x;&x\leq0\end{cases}\)

\(\frac{dy}{dx}=\begin{cases}1-2x;&x\geq0\\2x-1;&x\leq0\end{cases}\)

\(\frac{d^2y}{dx^2}=\begin{cases}-2;&x\geq0\\2;&x\leq0\end{cases}\)

Given:

y = |log_e x| \(\forall x>0\)

y = log_e x

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

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

Correct answer is D.

The surface area of a sphere, with radius r, is defined as

A(r) = 4πr² – (1)

Differentiating (1) with respect to t, we get

\(\frac{dA}{dt}=8\pi r\frac{dr}{dt}\)

= 8π×(current radius)×(rate of change of radius)

Total Surface Area of Cylinder = 2πr² + 2πrh

Given: Radius of the Cylinder varies.

Therefore, We need to find \(\frac{ds}{dr}\) where S = Surface Area of Cylinder and r = radius of Cylinder.

\(\frac{ds}{dr}=4\pi r + 2\pi h\)

Hence, Rate of change of total surface area of the cylinder when the radius is varying is given by (4πr + 2πh).

The volume of a Sphere \(=\frac{1}{6}\pi D^3\)

Where D = diameter of the Sphere

We need to find, \(\frac{dV}{dD}\) where V = Volume of the sphere and D = Diameter of the Sphere.

\(\frac{dV}{dD}=\frac{\pi D^2}{2}\)

Hence, Rate of change of Volume of Sphere with respect to the diameter of the Sphere is given by \(\frac{\pi D^2}{2}.\)