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Class 12 NCERT Solutions- Mathematics Part I – Chapter 5 Continuity And Differentiability – Exercise 5.2

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  • Last Updated : 03 Mar, 2021

 Differentiate the function with respect to x in Question 1 to 8

Question 1. Sin(x2 + 5)

Solution: 

y = sin(x2 + 5)

\frac{dy}{dx} = \frac{d(sin(x^2+5)}{dx}

= cos(x2 + 5) × \frac{d(x^2+5)}{dx}

= cos(x2 + 5) × (2x)

dy/dx = 2xcos(x2 + 5)

Question 2. cos(sin x)

Solution:

y = cos(sin x)

\frac{dy}{dx} = \frac{d \ cos(sin x)}{dx}

 = -sin(sin x) × \frac{d \ sinx}{dx}

= -sin(sin x)cos x  

Question 3. sin(ax + b)

Solution:

y = sin(ax + b)

\frac{dy}{dx}=\frac{d}{dx}sin(ax+b)

=cos(ax+b)\frac{d}{dy}(ax+b)

= a cos(ax + b)  

 Question 4. Sec(tan(√x)

Solution:

y = sec(tan√x)

\frac{dy}{dx} = \frac{d \ (sec(tan√x))  }{dx}    

= sec(tan √x) × tan(√x) × \frac{d \ (tan√x)  }{dx}

= sec (tan √x) × tan (tan √x) × sec2√x × \frac{d \ (√x)}{dx}    

= sec(tan√x)tan(tan√x)(sec2√x)1/(2√x)

= 1/(2√x) × sec(tan√x)tan(tan√x)(sec2√x)

Question 5. \frac{sin(ax+b)}{cos(cx+d)}

Solution:

y = \frac{sin(ax+b)}{cos(cx+d)}

\frac{dy}{dx} = \frac{d (\frac{sin(ax+b)}{cos(cx+d)})}{dx}

=\frac{cos(cx+d)\frac{d}{dx}sin(ax+b)-sin(ax+b)\frac{d}{dx}cos(cx+d)}{(cos(cx+d))}

=\frac{cos(cx+d)cos(ax+b)(a)+sin(ax+b)sin(cx+d)(c)}{cos^2(cx+b)}

Question 6. cos x3.sin2(x5)

Solution:

y = cos x3.sin2(x5)

\frac{dy}{dx} = cosx^3\frac{d(sin^2(x^5)}{dx}+sin^2(x^5)\frac{d(cosx^3)}{dx}

cosx^3.2sin(x^5)\frac{d(sin(x^5)}{dx}+sin^2(x^5)(-sinx^3)\frac{d(x^3)}{dx}

= cos x3.2sin(x5) .cos(x5(5x4)(5x4) – sin2(x5).sin x3.3x2

= 10x4 cos x3sin(x5)cos(x5) – 3x2 sin2(x5)sin x3

Question 7. 2√(cos(x2))

Solution:

y = 2√(cos(x2))

\frac{dy}{dx}   = \frac{\ d(2\sqrt{cos(x2})}{dx}

= 2\frac{\ d(\sqrt{cos(x^2})}{dx}

2\frac{1}{2\sqrt {cotx^2}}.\frac{d(cotx^2)}{dx}

\frac{1}{\sqrt {cot(x^2)}}.(-cosec^2(x^2)).\frac{d(x^2)}{dx}

\frac{1}{\sqrt {cot(x^2)}}.(-cosec^2(x^2)).2x

= \frac{-2xcosec^2x^2}{\sqrt{cot(x^2)}}

\frac{-2x}{sin^2(x^2).\sqrt{\frac{cos(x^2)}{sin(x^2)}}}

 \frac{-2x}{sin(x^2) \times sin(x^2)\sqrt{\frac{cos(x^2)}{sin(x^2)}}}

\frac{-2x}{sin(x^2) \times \sqrt{sin(x^2) \times \frac{cos(x^2)}{sin(x^2)}}}

\frac{-2x}{sin(x^2) \times \sqrt{cos(x^2) \times {sin(x^2)}}}

\frac{-2x}{sin(x^2) \times \sqrt{\frac{2}{2} \times cos(x^2) \times {sin(x^2)}}}

\frac{-2\sqrt{2}x}{sin(x^2) \times \sqrt{2 \times cos(x^2) \times {sin(x^2)}}}

\frac{-2\sqrt{2}x}{sin(x^2) \times \sqrt{sin(2x^2)}}

Question 8. cos (√x)

Solution:

y = cos (√x)

dy/dx = -sin√x\frac{d\sqrt{x}}{dx}

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

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

Question 9. Prove that the function f given by f(x) = |x – 1|, x ∈ R is not differentiable at x = 1.

Solution:

RHD=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}                                                                

\lim_{h\to 0}\frac{f(1+h)-f(1)}{h}                                                                 

= \lim_{h\to 0}\frac{((1+h)-1)-(1-1)}{h}      

\lim_{h\to 0}\frac{h-0}{h}   

=\lim_{h\to 0}(1)                                                        

= +1                                                                                                            

 LHD=\lim_{h\to 0}\frac{f(x)-f(x-h)}{h}     

\lim_{h\to 0}\frac{(1)-f(1-h)}{h}

\lim_{h\to 0}\frac{(1-1)-(-(1-h)-1)}{h}

\lim_{h\to 0}\frac{0-h}{h}

\lim_{h\to 0}(-1)

= -1   

LHD ≠ RHD  

Hence, f(x) is not differentiable at x = 1  

Question 10. Prove that the greatest integer function defined by f(x) = [x], 0 < x < 3 is not differentiable at x = 1 and x = 2.

Solution:

Given: f(x) = [x], 0 < x < 3

LHS:

f'(1) = \lim_{h\to0} \frac{f(x - h)-f(x)}{-h}

\lim_{h\to0} \frac{f(1 - h)-1}{-h}

=\lim_{h\to0} \frac{0-1}{-h}

= ∞

RHS:

f'(1) = \lim_{h\to0} \frac{f(x + h) - f(x)}{h}

\lim_{h\to0} \frac{f(1 + h) - f(1)}{h}

\lim_{h\to0} \frac{1-1}{h}

\lim_{h\to0} \frac{0}{h}

= 0

LHS ≠ RHS

So, the given f(x) = [x] is not differentiable at x = 1. 

Similarly, the given f(x) = [x] is not differentiable at x = 2. 


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