Differentiate the following functions with respect to x :

\displaystyle \text{Question 1: }  x^4 - 2 \sin x + 3 \cos x

Answer:

\displaystyle \frac{d}{dx} ( x^4 - 2 \sin x + 3 \cos x )

\displaystyle = \frac{d}{dx} (x^4) -2  \frac{d}{dx} ( \sin x ) + 3 \frac{d}{dx} (\cos x)

\displaystyle = 4 x^3 - 2 \cos x - 3 \sin x

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\displaystyle \text{Question 2: } 3^x + x^3 + 3^3 

Answer:

\displaystyle \frac{d}{dx} (3^x + x^3 + 3^3)

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

\displaystyle = 3^x \log 3 + 3 x^2

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\displaystyle \text{Question 3: }  \frac{x^3}{3} - 2 \sqrt{x} + \frac{5}{x^2}

Answer:

\displaystyle \frac{d}{dx} (\frac{x^3}{3} - 2 \sqrt{x} + \frac{5}{x^2})

\displaystyle = \frac{d}{dx} (\frac{x^3}{3}) -2 \frac{d}{dx} (\sqrt{x}) + \frac{d}{dx} (\frac{5}{x^2})

\displaystyle = \frac{1}{3} ( 3 x^2) - 2  \cdot \frac{1}{2} x^{\frac{-1}{2}} + 5 ( -2) x^{-3}

\displaystyle = x^2 - x^{\frac{-1}{2}} - 10 x^{-3}

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\displaystyle \text{Question 4: } e^{x \log a} + e^{a \log x} + e^{ a \log a} 

Answer:

\displaystyle \frac{d}{dx} (e^{x \log a} + e^{a \log x} + e^{ a \log a})

\displaystyle = \frac{d}{dx} (e^{x \log a}) + \frac{d}{dx} (e^{a \log x}) + \frac{d}{dx} (e^{ a \log a})

\displaystyle = \frac{d}{dx} (a^x) + \frac{d}{dx} (x^a) + \frac{d}{dx} (a^a)

\displaystyle = a^x \log a + a x^{a-1}

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\displaystyle \text{Question 5: }  (2x^2+1)(3x+2)

Answer:

\displaystyle \frac{d}{dx} ( (2x^2+1)(3x+2))

\displaystyle \frac{d}{dx} ( 6x^3+4x^2+3x+2)

\displaystyle = \frac{d}{dx} (6x^3) + \frac{d}{dx} (4x^2) + \frac{d}{dx} (3x) + \frac{d}{dx} (2)

\displaystyle = 18x^2+8x+3

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\displaystyle \text{Question 6: }  \log_3 x + 3 \log_e x+ 2 \tan x

Answer:

\displaystyle \frac{d}{dx} (\log_3 x + 3 \log_e x+ 2 \tan x)

\displaystyle = \frac{d}{dx} (\log_3 x) + 3 \frac{d}{dx} (\log_e x) + 2 \frac{d}{dx} ( \tan x)

\displaystyle = \frac{d}{dx} (\frac{\log x}{\log 3}) + 3 \frac{d}{dx} (\log_e x) + 2 \frac{d}{dx} ( \tan x)

\displaystyle = \frac{1}{\log 3} \cdot \frac{1}{x} + 3 \frac{1}{x} + 2 \sec^2 x

\displaystyle = \frac{1}{x \log 3} + \frac{3}{x} + 2 \sec^2 x

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\displaystyle \text{Question 7: }  \Big(x + \frac{1}{x} \Big) \Big( \sqrt{x}+ \frac{1}{\sqrt{x}} \Big)

Answer:

\displaystyle \frac{d}{dx} \Big(x + \frac{1}{x} \Big) \Big( \sqrt{x}+ \frac{1}{\sqrt{x}} \Big)

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

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

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

\displaystyle = \frac{3}{2} x^{\frac{1}{2}} + \frac{1}{2} x^{\frac{-1}{2}} - \frac{1}{2} x^{\frac{-3}{2}} - \frac{3}{2} x^{\frac{-5}{2}}

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\displaystyle \text{Question 8: }  \Big(\sqrt{x}+ \frac{1}{\sqrt{x}} \Big)^3

Answer:

\displaystyle \frac{d}{dx} (\Big(\sqrt{x}+ \frac{1}{\sqrt{x}} \Big)^3)

\displaystyle =\frac{d}{dx} \Big(  (\sqrt{x})^3+ 3 ( \sqrt{x})^2 \Big( \frac{1}{\sqrt{x} } \Big) + 3 (\sqrt{x}) \Big(\frac{1}{\sqrt{x}}\Big)^2 + \Big( \frac{1}{\sqrt{x}} \Big)^3    \Big)

\displaystyle = \frac{d}{dx} \Big(x^{\frac{3}{2} } \Big) + 3 \frac{d}{dx} \Big(x^{\frac{1}{2}}\Big) + 3\frac{d}{dx} \Big(x^{\frac{-1}{2}} \Big) + \frac{d}{dx} \Big( x^{\frac{-3}{2}} \Big)

\displaystyle = \frac{3}{2} x^{\frac{1}{2}} + 3 \cdot \frac{1}{2} x^{\frac{-1}{2}} + 3 \cdot \frac{-1}{2} x^{ \frac{-3}{2}} - \frac{3}{2} x^{\frac{-5}{2}}

\displaystyle = \frac{3}{2} x^{\frac{1}{2}} +  \frac{3}{2} x^{\frac{-1}{2}} - \frac{3}{2} x^{ \frac{-3}{2}} - \frac{3}{2} x^{\frac{-5}{2}}   

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\displaystyle \text{Question 9: } \frac{2x^2+ 3x+ 4}{x} 

Answer:

\displaystyle \frac{d}{dx} (\frac{2x^2+ 3x+ 4}{x})

\displaystyle = \frac{d}{dx} (\frac{2x^2}{x}) + \frac{d}{dx} (\frac{3x}{x}) + \frac{d}{dx} (\frac{4}{x})

\displaystyle = 2\frac{d}{dx} (x) + 3\frac{d}{dx} (1) + 4\frac{d}{dx} (x^{-1})

\displaystyle = 2 + 3(0) + 4 ( -1) x^{-2}

\displaystyle = 2 - \frac{4}{x^2}

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\displaystyle \text{Question 10: } \frac{(x^3+1)(x-2)}{x^2} 

Answer:

\displaystyle \frac{d}{dx} (\frac{(x^3+1)(x-2)}{x^2})

\displaystyle = \frac{d}{dx} (\frac{x^4 - 2 x^3 + x - 2}{x^2})

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

\displaystyle = 2x - 2 - x^{-2} -2 ( -2) x^{-3}

\displaystyle = 2x - 2 - \frac{1}{x^2}  + \frac{4}{x^3}

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\displaystyle \text{Question 11: } \frac{a \cos x + b \sin x + c }{\sin x} 

Answer:

\displaystyle \frac{d}{dx} (\frac{a \cos x + b \sin x + c }{\sin x})

\displaystyle = \frac{d}{dx} \Big(\frac{a \cos x}{\sin x} \Big) + \frac{d}{dx} \Big(\frac{b \sin x}{\sin x} \Big) + \frac{d}{dx} \Big(\frac{c}{\sin x} \Big)

\displaystyle = a \frac{d}{dx} cot x + \frac{d}{dx} \Big(\frac{b \sin x}{\sin x} \Big) + \frac{d}{dx} \Big(\frac{c}{\sin x} \Big)

\displaystyle = a \frac{d}{dx} (\cot x) + \frac{d}{dx} (b) + c\frac{d}{dx} (\mathrm{cosec} x)

\displaystyle = - a \mathrm{cosec}^2 x + 0 - c \mathrm{cosec} x \cdot \cot x 

\displaystyle = - a \mathrm{cosec}^2 x - c \mathrm{cosec} x \cdot \cot x 

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\displaystyle \text{Question 12: }  2 \sec x + 3 \cot x- 4 \tan x

Answer:

\displaystyle \frac{d}{dx} (2 \sec x + 3 \cot x- 4 \tan x)

\displaystyle = 2\frac{d}{dx} (\sec x) + 3\frac{d}{dx} (\cot x) -4 \frac{d}{dx} (\tan x)

\displaystyle = 2 \sec x \tan x - 3 \mathrm{cosec}^2 x - 4 \sec^2 x

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\displaystyle \text{Question 13: } a_0 x^n + a_1 x^{n-1} + a_2 x^{n-2} + \cdots + a_{n-1}x + a_n 

Answer:

\displaystyle \frac{d}{dx} (a_0 x^n + a_1 x^{n-1} + a_2 x^{n-2} + \cdots + a_{n-1}x + a_n)

\displaystyle = a_0 \frac{d}{dx} (x^n) + a_1\frac{d}{dx} (x^{n-1}) +a_2 \frac{d}{dx} (x^{n-2}) + \cdots + a_{n-1} \frac{d}{dx} (x) + a_n\frac{d}{dx} (1)

\displaystyle = na_0 x^{n-1} +(n-1) a_1 x^{n-2} + ( n-2) a_2 x^{n-3} + \cdots + a_{n-1}

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\displaystyle \text{Question 14: } \frac{1}{\sin x} + 2^{x+3} + \frac{4}{\log_x 3} 

Answer:

\displaystyle \frac{d}{dx} \Big(\frac{1}{\sin x} +2^{x+3}  + \frac{4}{\log_x 3} \Big)

\displaystyle = \frac{d}{dx} \Big( \mathrm{cosec} x  + 2^3 \cdot 2^x+ \frac{4}{\frac{\log 3}{\log x}} \Big)

\displaystyle = \frac{d}{dx} ( \mathrm{cosec} x) + 2^3\frac{d}{dx} (2^x) + \frac{4}{\log 3} \frac{d}{dx} (\log x)

\displaystyle = - \mathrm{cosec} x \cot x + 2^3 \cdot 2^x \cdot \log 2  + \frac{4}{\log 3} \cdot \frac{1}{x}

\displaystyle = - \mathrm{cosec} x \cot x + 2^{x+3} \cdot \log 2 + \frac{4}{x\log 3}

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\displaystyle \text{Question 15: } \frac{(x+5)(2x^2-1)}{x} 

Answer:

\displaystyle \frac{d}{dx} \Big(\frac{(x+5)(2x^2-1)}{x} \Big)

\displaystyle = \frac{d}{dx} \Big(\frac{2x^3+10x^2-x-5}{x} \Big)

\displaystyle = \frac{d}{dx} (2x^2) + \frac{d}{dx} (10x) - \frac{d}{dx} (1) - \frac{d}{dx} (5x^{-1})

\displaystyle = 4x + 10 + 5x^{-2}

\displaystyle = 4x + 10 + \frac{5}{x^2}

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\displaystyle \text{Question 16: } \log \Bigg( \frac{1}{\sqrt{x}} \Bigg) + 5x^a - 3a^x + \sqrt[3]{x^2}+ 6 \sqrt[4]{x^{-3}}

Answer:

\displaystyle \frac{d}{dx} \Bigg[ \log \Bigg( \frac{1}{\sqrt{x}} \Bigg) + 5x^a - 3a^x + \sqrt[3]{x^2}+ 6 \sqrt[4]{x^{-3}} \Bigg]

\displaystyle = \frac{d}{dx} \Big( \log ( x^{\frac{-1}{2}} \Big) + 5\frac{d}{dx} (x^a) -3 \frac{d}{dx} (a^x) + \frac{d}{dx} (x^{\frac{2}{3}}) + 6\frac{d}{dx} (x^{\frac{-3}{4}})  

\displaystyle = \frac{d}{dx} \Big( \frac{-1}{2} \log x \Big) + 5\frac{d}{dx} (x^a) -3 \frac{d}{dx} (a^x) + \frac{d}{dx} (x^{\frac{2}{3}}) + 6\frac{d}{dx} (x^{\frac{-3}{4}})  

\displaystyle = \frac{-1}{2} \cdot \frac{1}{x} + 5a x^{a-1} - 3a^x \log a + \frac{2}{3} x^{\frac{-1}{3}} + 6 \Big( \frac{-3}{4} \Big) x^{\frac{-7}{4}}

\displaystyle = \frac{-1}{2x}  + 5a x^{a-1} - 3a^x \log a + \frac{2}{3} x^{\frac{-1}{3}} - \frac{9}{2} x^{\frac{-7}{4}}

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\displaystyle \text{Question 17: }  \cos ( x+a) 

Answer:

\displaystyle \frac{d}{dx} (\cos ( x+a) )

\displaystyle = \frac{d}{dx} (\cos x \cos a - \sin x \sin a )

\displaystyle = \cos a\frac{d}{dx} (\cos x) - \sin a \frac{d}{dx} (\sin x) 

\displaystyle = - \cos a \sin x - \sin a \cos x

\displaystyle = - ( \sin x \cos a + \cos x \sin a )

\displaystyle = - \sin ( x+ a)

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\displaystyle \text{Question 18: } \frac{\cos ( x-2)}{\sin x} 

Answer:

\displaystyle \frac{d}{dx} \Big( \frac{\cos ( x-2)}{\sin x} \Big)

\displaystyle = \frac{d}{dx} \Big( \frac{\cos x \cos 2 + \sin x \sin 2}{\sin x} \Big)

\displaystyle = \cos 2 \frac{d}{dx} (\cot x) + \sin 2\frac{d}{dx} (1) 

\displaystyle = - \cos 2 \ \mathrm{cosec}^2 x + 0 

\displaystyle = - \cos 2 \ \mathrm{cosec}^2 x

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\displaystyle \text{Question 19: } \text{If } y = \Bigg( \sin \frac{x}{2} + \cos \frac{x}{2} \Bigg)^2, \text{ find } \frac{dy}{dx} \text{ at } x = \frac{\pi}{4}

Answer:

\displaystyle \frac{d}{dx} \Bigg( \sin \frac{x}{2} + \cos \frac{x}{2} \Bigg)^2

\displaystyle = \frac{d}{dx} \Bigg( \sin^2 \frac{x}{2} + \cos^2 \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2} \Bigg)

\displaystyle = \frac{d}{dx} ( 1 + \sin x)

\displaystyle = 0 + \cos x

\displaystyle = \cos x

\displaystyle \frac{dy}{dx} \text{ at } x = \frac{\pi}{6} = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}

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\displaystyle \text{Question 20: }  \text{If } y = \Bigg( \frac{2 - 3 \cos x}{\sin x} \Bigg) , \text{ find } \frac{dy}{dx} \text{ at } x = \frac{\pi}{4}

Answer:

\displaystyle \frac{d}{dx} (\Bigg( \frac{2 - 3 \cos x}{\sin x} \Bigg))

\displaystyle = \frac{d}{dx} ( 2 \mathrm{cosec} x - 3 \cot x)

\displaystyle = 2\frac{d}{dx} ( \mathrm{cosec} x) -3 \frac{d}{dx} (\cot x) 

\displaystyle = - 2 \mathrm{cosec} x \cdot \cot x + 3 \mathrm{cosec}^2 x

\displaystyle \frac{dy}{dx} \text{ at } x = \frac{\pi}{4} = - 2 \mathrm{cosec} \frac{\pi}{4} \cdot \cot \frac{\pi}{4} + 3 \mathrm{cosec}^2 \frac{\pi}{4} \\ \\ { \hspace{2.5cm} = -2 \sqrt{2} \cdot 1 + 3 \cdot 2 = - 2\sqrt{2} + 6 = 6 - 2\sqrt{2}}

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\displaystyle \text{Question 21: }  \text{Find the slope of the tangent to the curve } \\ \\ f(x) = 2x^6 + x^4 - 1 \text{ at } x = 1

Answer:

\displaystyle \frac{d}{dx} (2x^6 + x^4 - 1)

\displaystyle = \frac{d}{dx} (2x^6) + \frac{d}{dx} (x^4) - \frac{d}{dx} (1)

\displaystyle = 12x^5 + 4 x^3

\displaystyle \frac{dy}{dx} \text{ at } x =1 = 12(1)^5 + 4(1)^3 = 12 + 4 = 16

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\displaystyle \text{Question 22: } \text{If } y = \sqrt{\frac{x}{a}} + \sqrt{\frac{a}{x}}, \text{ prove that } 2xy \frac{dy}{dx} = \Bigg( \frac{x}{a} - \frac{a}{x} \Bigg)   

Answer:

\displaystyle \frac{dy}{dx} = \frac{d}{dx} \Big( \sqrt{\frac{x}{a}} + \sqrt{\frac{a}{x}} \Big)

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

\displaystyle \frac{dy}{dx} = \frac{1}{\sqrt{a}} \frac{1}{2\sqrt{x}} + \sqrt{a} \cdot \frac{-1}{2} \cdot  \frac{1}{x\sqrt{x}}

\displaystyle \frac{dy}{dx} = \frac{1}{2x} \Bigg(  \sqrt{ \frac{x}{a} } -  \sqrt{ \frac{a}{x} } \Bigg)   

\displaystyle 2x\frac{dy}{dx} =  \Bigg(  \sqrt{ \frac{x}{a} } -  \sqrt{ \frac{a}{x} } \Bigg)   

Multiplying both sides by y we get

\displaystyle 2xy\frac{dy}{dx} =  \Bigg(  \sqrt{ \frac{x}{a} } -  \sqrt{ \frac{a}{x} } \Bigg) \Bigg(  \sqrt{ \frac{x}{a} } +  \sqrt{ \frac{a}{x} } \Bigg)   

\displaystyle 2xy\frac{dy}{dx} = \Bigg( \frac{x}{a} - \frac{a}{x} \Bigg)  

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\displaystyle \text{Question 23: }  \text{Find the rate at which the function } \\ \\ f(x) = x^4 - 2x^3 + 3x^2 + x + 5  \text{ changes with respect to } x   

Answer:

\displaystyle \text{Rate } = f'(x)

\displaystyle = \frac{d}{dx} (x^4 - 2x^3 + 3x^2 + x + 5)

\displaystyle = \frac{d}{dx} (x^4) -2 \frac{d}{dx} (x^3) +3 \frac{d}{dx} (x^2) + \frac{d}{dx} (5)

\displaystyle = 4x^3- 6x^2 + 6x + 1

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\displaystyle \text{Question 24: }  \text{If } y = \frac{2x^9}{3} - \frac{5}{7} x^7 + 6x^3 - x , \text{ find } \frac{dy}{dx} \text{ at } x = 1

Answer:

\displaystyle \frac{dy}{dx} = \frac{d}{dx} \Big( \frac{2x^9}{3} - \frac{5}{7} x^7 + 6x^3 - x \Big)

\displaystyle = \frac{2}{3}\frac{d}{dx} (x^9) -\frac{5}{7}  \frac{d}{dx} (x^7) + 6\frac{d}{dx} (x^3) - \frac{d}{dx} (x)

\displaystyle = \frac{2}{3} ( 9x^8) -\frac{5}{7}  (7x^6) + 6 (3x^2) - 1

\displaystyle = 6x^8 - 5x^6 + 18x^2 - 1

\displaystyle \frac{dy}{dx} \text{ at } x =1 = 6(1)^8 - 5(1)^6 + 18(1)^2 - 1  = 6-5+18-1 = 18

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\displaystyle \text{Question 25: }  \text{If for } f(x) = \lambda x^2 + \mu x + 12, f'(4) = 15 \text{ and } f'(2) = 11 , \text{ then find} \lambda \text{ and } \mu .

Answer:

\displaystyle f'(x) = \frac{d}{dx} (\lambda x^2 + \mu x + 12)

\displaystyle = \lambda \frac{d}{dx} (x^2) + \mu \frac{d}{dx} (x) + \frac{d}{dx} (12)

\displaystyle = 2 \lambda  x + \mu

Given \displaystyle f'(4) = 15

\displaystyle \Rightarrow 2 \lambda ( 4) + \mu = 15

\displaystyle \Rightarrow 8 \lambda + u = 15   \text{    ... ... ... ... ... ... (i) }

Also \displaystyle f'(2) = 11

\displaystyle \Rightarrow 2 \lambda ( 2) + \mu = 11

\displaystyle \Rightarrow 4 \lambda + u = 11   \text{    ... ... ... ... ... ... (ii) }

Solving (i) and (ii) we get \displaystyle \lambda = 1 and \displaystyle \mu = 7

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\displaystyle \text{Question 26: }  \text{For the function } f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + \cdots + \frac{x^2}{2}+ x + 1. \\ \\ \text{Prove that } f'(1) = 100 f'(0) . 

Answer:

\displaystyle f'(x) = \frac{d}{dx} (\frac{x^{100}}{100} + \frac{x^{99}}{99} + \cdots + \frac{x^2}{2}+ x + 1)

\displaystyle \Rightarrow f'(x)= x^{99} + x^{98} + \cdots + x + 1

\displaystyle \Rightarrow f'(1)= 1^{99} + 1^{98} + \cdots + 1 + 1 = 99+1 = 100

\displaystyle \Rightarrow f'(0)= 0^{99} + 0^{98} + \cdots + 0 + 1 = 0+1 = 1

Therefore we see that \displaystyle f'(1) = 100 f'(0) . 

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