Differentiation is a fundamental concept in calculus that deals with the rate at which a function changes at any given point. It is used to find the slope of the curve which is represented by the derivative of the function. Differentiation has applications in various fields such as physics, engineering, economics, and biology. In this section, we will provide detailed solutions to Exercise 11.7 (Set 2) from RD Sharma's Class 12 Mathematics textbook.
Differentiation
Differentiation is the mathematical process of finding the rate at which a function changes at any given point. It involves calculating the derivative in which represents the slope of the function's graph at a specific point. Differentiation helps determine the instantaneous rate of change and is essential in various applications such as physics, engineering, and economics. The notation for the derivative of the function f(x) with the respect to the x is
Question 11. Find\frac{dy}{dx} , when x=\frac{2t}{1+t^2} andy=\frac{1-t^2}{1+t^2}
Solution:
Here,
x=\frac{2t}{1+t^2} Differentiating it with respect to t using quotient rule,
\frac{dx}{dt}=\left[\frac{(1+t^2)\frac{d}{dt}(2t)-2t\frac{d}{dt}(1+t^2)}{(1+t^2)^2}\right]\\ =\left[\frac{(1+t^2)(2)-2t(2t)}{(1+t^2)^2}\right]\\ =\left[\frac{2+2t^2-4t^2}{(1+t^2)^2}\right]\\ =\left[\frac{2+2t^2-4t^2}{(1+t^2)^2}\right]\\ =\left[\frac{2-2t^2}{(1+t^2)^2}\right]\\ \frac{dx}{dt}=\frac{2(1-t^2)}{(1+t^2)^2}\ \ \ \ \ ....(1) and,
y=\frac{1-t^2}{1+t^2} Differentiating it with respect to t using quotient rule,
\frac{dy}{dt}=\left[\frac{(1+t^2)\frac{d}{dt}(1-t^2)-(1-t^2)\frac{d}{dt}(1+t^2)}{(1+t^2)^2}\right]\\ =\left[\frac{(1+t^2)(-2t)-(1-t^2)(2t)}{(1+t^2)^2}\right]\\ =\left[\frac{-2t-2t^3-2t+2t^3}{(1+t^2)^2}\right]\\ \frac{dy}{dt}=\frac{-4t}{(1+t^2)^2}\ \ \ \ \ ....(2) Dividing equation (2) by (1)
\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-4t}{(1+t^2)^2}\times\frac{(1+t^2)^2}{2(1-t^2)}\\ =\frac{-2t}{1-t^2}\\ \frac{dy}{dx}=-\frac{x}{y}\ \ \ \ \left[Since,\ \frac{x}{y}=\frac{2t}{1+t^2}\times\frac{1+t^2}{1-t^2}=\frac{2t}{1-t^2}\right]
Question 12. Find\frac{dy}{dx} , when x=cos^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right) andy=sin^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right)
Solution:
Here,
x=cos^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right) Differentiating it with respect to t using chain rule,
\frac{dx}{dt}=\frac{-1}{\sqrt{1-\left(\frac{1}{1+t^2}\right)}^2}\frac{d}{dt}\left(\frac{1}{\sqrt{1+t^2}}\right)\\ =\frac{-1}{\sqrt{1-\frac{1}{(1+t^2)}}}\left[\frac{-1}{2(1+t^2)^{\frac{3}{2}}}\right]\frac{d}{dt}(1+t^2)\\ =\frac{(1+t^2)^{\frac{1}{2}}}{\sqrt{1+t^2-1}}\times\frac{-1}{2(1+t^2)^{\frac{3}{2}}}(2t)\\ =\frac{-t}{\sqrt{t^2}\times(1+t^2)}\\ \frac{dx}{dt}=\frac{-1}{1+t^2}\ \ \ \ \ ....(1) Now,
y=sin^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right) Differentiating it with respect to t using chain rule,
\frac{dy}{dt}=\frac{1}{\sqrt{1-\frac{1}{(\sqrt{1+t^2})^2}}}\times\frac{d}{dt}\left(\frac{1}{\sqrt{1+t^2}}\right)\ \ \ \ .....(2) Dividing equation (2) by (1)
\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{1}{(1+t^2)}\times\frac{(1+t^2)}{-1}\\ \frac{dy}{dx}=1
Question 13. Find\frac{dy}{dx} , when x=\frac{1-t^2}{1+t^2} andy=\frac{2t}{1+t^2}
Solution:
Here,
x=\frac{1-t^2}{1+t^2} Differentiating it with respect to t using quotient rule,
\frac{dx}{dt}=\left[\frac{(1+t^2)\frac{d}{dt}(1-t^2)-(1-t^2)\frac{d}{dt}(1+t^2)}{(1+t^2)^2}\right]\\ =\left[\frac{(1+t^2)(-2t)-(1-t^2)(2t)}{(1+t^2)^2}\right]\\ =\left[\frac{-2t-2t^3-2t+2t^3}{(1+t^2)^2}\right]\\ \frac{dx}{dt}=\frac{-4t}{(1+t^2)^2}\ \ \ \ \ ....(1) and,
y=\frac{2t}{1+t^2} Differentiating it with respect to t using quotient rule,
\frac{dy}{dt}=\left[\frac{(1+t^2)\frac{d}{dt}(2t)-2t\frac{d}{dt}(1+t^2)}{(1+t^2)^2}\right]\\ =\left[\frac{(1+t^2)(2)-2t(2t)}{(1+t^2)^2}\right]\\ =\left[\frac{2+2t^2-4t^2}{(1+t^2)^2}\right]\\ \frac{dx}{dt}=\frac{2(1-t^2)}{(1+t^2)^2}\ \ \ \ \ ....(2)
Question 14. If x = 2cosθ - cos2θ and y = 2sinθ - sin2θ, prove that\frac{dy}{dx}=tan\left(\frac{3\theta}{2}\right)
Solution:
Here,
x = 2cosθ - cos2θ
Differentiating it with respect to θ using chain rule,
\frac{dx}{dθ}=2(-sinθ )-(-sin2θ)\frac{d}{dθ} (2θ )\\ =-2sinθ +2sin2θ \\ \frac{dx}{dθ}=2(sin2θ -sinθ )\ \ \ \ \ ....(1) and,
y = 2sinθ - sin2θ
Differentiating it with respect to θ using chain rule,
\frac{dy}{dθ}=2cosθ -cos2θ \frac{d}{dθ}(2θ )\\ =2cosθ -cos2θ (2)\\ =2cosθ -2cos2θ \\ \frac{dy}{dθ}=2(cosθ -cos2θ )\ \ \ \ \ ....(2) Dividing equation (2) by equation (1),
\frac{\frac{dy}{dθ}}{\frac{dx}{dθ}}=\frac{2(cosθ -cos2θ )}{2(sin2θ - sinθ )}\\ =\frac{cosθ -cos2θ}{sin2θ -sinθ}\\ \frac{dy}{dx}=\frac{-2sin\left(\frac{θ +2θ}{2}\right)sin\left(\frac{θ -2θ}{2}\right)}{2cos\left(\frac{2θ +θ}{2}\right)sin\left(\frac{2θ -θ}{2}\right)}\ \ \ \ \left[Since,\ sinA-sinB=2cos\left(\frac{A+B}{2}\right)sin\left(\frac{A-B}{2}\right),\ \ cosA-cosB=-2sin\left(\frac{A+B}{2}\right)sin\left(\frac{A-B}{2}\right)\right]\\ =\frac{-sin\left(\frac{3θ )}{2}\right)\left(sin\left(\frac{-θ}{2}\right)\right)}{cos\left(\frac{3θ}{2}\right)sin\left(\frac{θ}{2}\right)}\\ =\frac{sin\left(\frac{3θ}{2}\right)}{cos\left(\frac{3θ}{2}\right)}\\ \frac{dy}{dx}=tan\left(\frac{3θ}{2}\right)
Question 15. If x = ecos2t and y = esin2t prove that,\frac{dy}{dx}=\frac{y\ log\ x}{x\ log\ y}
Solution:
Here,
x = ecos2t
Differentiating it with respect to t using chain rule,
\frac{dx}{dt}=\frac{d}{dt}(e^{cos2t})\\ =e^{cos2t}\frac{d}{dt}(cos2t)\\ =e^{cos2t}(-sin2t)\frac{d}{dt}(2t)\\ \frac{dx}{dt}=-sin2te^{cos2t}\ \ \ \ \ .....(1) and,
y = esin2t
Differentiating it with respect to t using chain rule,
\frac{dy}{dt}=\frac{d}{dt}(e^{sin2t})\\ =e^{sin2t}\frac{d}{dt}(sin2t)\\ =e^{sin2t}(cos2t)\frac{d}{dt}(2t)\\ =e^{sin2t}(cos2t)(2)\\ \frac{dy}{dt}=2cos2te^{sin2t}\ \ \ \ \ .....(2) Dividing equation (2) by (1)
\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2cos2te^{sin2t}}{-2sin2te^{cos2t}}\\ \frac{dy}{dx}=-\frac{ylogx}{xlogy}
\left[Since,\ x=e^{cos2t}\Rightarrow logx=cos2t\ \ y=e^{sin2t}\Rightarrow logy=sin2t\right]
Question 16. If x = cos t and y = sin t, prove that\frac{dy}{dx}=\frac{1}{\sqrt3}\ at\ t=\frac{2x}{3}
Solution:
Here,
x = cos t
Differentiating it with respect to t,
\frac{dx}{dt}=\frac{d}{dt}(cos\ t)\\ \frac{dx}{dt}=-sin\ t\ \ \ \ \ ....(1) and,
y = sin t
Differentiating it with respect to t,
\frac{dx}{dt}=\frac{d}{dt}(sin\ t)\\ \frac{dx}{dt}=cos\ t\ \ \ \ \ ....(2) Dividing equation (2) by (1),
\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{cos\ t}{-sin\ t}\\ \frac{dy}{dx}=-cot\ t\\ \left(\frac{dy}{dx}\right)=-cot\left(\frac{2\pi}{3}\right)\\ =-cot\left(\pi-\frac{\pi}{3}\right)\\ =-\left[-cot\left(\frac{\pi}{3}\right)\right]\\ =cot\left(\frac{\pi}{3}\right)\\ \frac{dy}{dx}=\frac{1}{\sqrt3}
Question 17. Ifx=a\left(t+\frac{1}{t}\right) and y=a\left(t-\frac{1}{t}\right) , Prove that\frac{dy}{dx}=\frac{x}{y}
Solution:
Here,
x=a\left(t+\frac{1}{t}\right) Differentiating it with respect to t,
\frac{dx}{dt}=a\frac{d}{dt}\left(t+\frac{1}{t}\right)\\ =a\left(1-\frac{1}{t^2}\right)\\ \frac{dx}{dt}=a\left(\frac{t^2-1}{t^2}\right)\ \ \ \ \ .....(1) and,
y=a\left(t-\frac{1}{t}\right) Differentiating it with respect to t,
\frac{dy}{dt}=a\frac{d}{dt}\left(t-\frac{1}{t}\right)\\ =a\left(1+\frac{1}{t^2}\right)\\ \frac{dy}{dt}=a\left(\frac{t^2+1}{t^2}\right)\ \ \ \ \ .....(2) Dividing equation (2) by (1)
\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=a\frac{(t^2+1)}{t^2}\times\frac{t^2}{a(t^2-1)}\\ \frac{dy}{dx}=\frac{t^2+1}{t^2-1}\\ \frac{dy}{dx}=\frac{x}{y}\ \ \ \ \ \ \left[Since,\ \frac{x}{y}=\frac{a(t^2+1)}{t}\times\frac{t}{a(t^2-1)}=\left(\frac{t^2+1}{t^2-1}\right)\right]
Question 18. Ifx=sin^{-1}\left(\frac{2t}{1+t^2}\right) andy =tan^{-1}\left(\frac{2t}{1-t^2}\right) , -1 < 1 < 1, prove that\frac{dy}{dx}=1
Solution:
Here,
x=sin^{-1}\left(\frac{2t}{1+t^2}\right) Put t = tan θ
x=sin^{-1}\left(\frac{2tanθ}{1+tan^2θ}\right)\\ =sin^{-1}(sin2θ )\\ =2θ \ \ \ \ \ \left[Since,\ sin\ 2x=\frac{2tan\ x}{1+tan^2x}\right]\\ x=2(tan^{-1}t)\ \ \ \ \ [Since,\ t=sin\ θ ] Differentiating it with respect to t,
\frac{dx}{dt}=\frac{2}{1+t^2}\ \ \ \ \ .....(1) Further,
y =tan^{-1}\left(\frac{2t}{1-t^2}\right) Put t = tan θ
y=tan^{-1}\left(\frac{2tanθ}{1+tan^2θ}\right)\\ =tan^{-1}(tan2θ )\\ =2θ \ \ \ \ \ \left[Since,\ tan\ 2x=\frac{2tan\ x}{1-tan^2x}\right]\\ y=2tan^{-1}t\ \ \ \ \ [Since,\ t=tan\ θ ] Differentiating it with respect to t,
\frac{dy}{dt}=\frac{2}{1+t^2}\ \ \ \ \ \ ......(2) Dividing equation (2) by (1),
\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2}{1+t^2}\times\frac{1+t^2}{2}\\ \frac{dy}{dx}=1
Question 19. If x and y are connected parametrically by the equation, without eliminating the parameter, find\frac{dy}{dx} , when: x=\frac{sin^3t}{\sqrt{cos2t}} ,y=\frac{cos^3t}{\sqrt{cos2t}}
Solution:
Here, the given equations are
x=\frac{sin^3t}{\sqrt{cos2t}} andy=\frac{cos^3t}{\sqrt{cos2t}} Thus,
\frac{dx}{dt}\frac{d}{dt}\left[\frac{sin^3t}{\sqrt{cos2t}}\right]\\ =\frac{\sqrt{cos2t}.\frac{d}{dt}(sin^3t)-sin^3t.\frac{d}{dt}\sqrt{cos2t}}{cos2t}\\ =\frac{\sqrt{cos2t}.3sin^2t.\frac{d}{dt}(sin\ t)-sin^3t\times\frac{1}{2\sqrt{cos\ 2t}}.\frac{d}{dt}(cos2t)}{cos2t}\\ =\frac{3\sqrt{cos\ 2t}.sin^2t\ cos\ t-\frac{sin^3t}{2\sqrt{cos\ 2t}}.(-2sin\ 2t)}{cos\ 2t}\\ =\frac{3cos\ 2t\ sin^2tcos\ t+sin^3tsin\ 2t}{cos\ 2t\sqrt{cos\ 2t}}\\ \frac{dy}{dt}=\frac{d}{dt}\left[\frac{cos^3t}{\sqrt{cos\ 2t}}\right]\\ =\frac{\sqrt{cos\ 2t}.\frac{d}{dt}(cos^3t)-cos^3t.\frac{d}{dt}(\sqrt{cos\ 2t})}{cos\ 2t}\\ =\frac{-3cos\ 2t.cos^2t.sin\ t+cos^3t\ sin\ 2t}{cos\ 2t.\sqrt{cos\ 2t}} Therefore,
\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-3cos\ 2t.cos^2t. sin\ t+cos^3t\ sin\ 2t}{3cos\ 2t\ sin^2t\ cos\ t+sin^3t\ sin\ 2t}\\ =\frac{-3cos\ 2t\ cos^2t.sin\ t+cos^3t(2sin\ t\ cos\ t)}{3cos\ 2t\ sin^2t\ cos\ t+sin^3t(2sin\ t+2sin^3t)}\\ =\frac{[-3(2cos^2t-1)cos\ t+2cos^3t]}{[3(1-2sin^2t)sin\ t+2sin^3t]}\ \ \ \ \ \ \ [cos\ 2t=(2cos^2t-1),\ cos\ 2t=(1-2sin^2t)]\\ =\frac{-4cos^3t+3cos\ t}{3sin\ t-4sin^3\ t}\\ =\frac{-cos\ 3t}{sin\ 3t}\ \ \ \ \ \ [cos\ 3t=4cos^3t3cos\ t,\ sin\ 3t=3sin\ t-4sin^3t]\\ =-cot3t
Question 20. Ifx=\left(t+\frac{1}{t}\right)^a andy=a^{\left(t+\frac{1}{t}\right)} , find\frac{dy}{dx}
Solution:
Here,
x=\left(t+\frac{1}{t}\right)^a Differentiating it with respect to t using chain rule,
\frac{dx}{dt}=\frac{d}{dt}\left(t+\frac{1}{t}\right)^a\\ =a\left(t+\frac{1}{t}\right)^{a-1}\frac{d}{dt}\left(t+\frac{1}{t}\right)\\ \frac{dx}{dt}=a\left(t+\frac{1}{t}\right)^{1-1}\left(1-\frac{1}{t^2}\right)\ \ \ \ \ .....(1) And,
y=a^{\left(t+\frac{1}{t}\right)} Differentiating it with respect to t using chain rule,
\frac{dy}{dt}=\frac{d}{dt}a^{\left(t+\frac{1}{t}\right)}\\ =a^{\left(t+\frac{1}{t}\right)}\times log\ a\ \frac{d}{dt}\left(t+\frac{1}{t}\right)\\ \frac{dy}{dt}=a^{\left(t+\frac{1}{t}\right)}\times log\ a\left(1-\frac{1}{t^2}\right)\ \ \ \ \ ......(2) Dividing equation (2) by (1)
\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{a^{\left(t+\frac{1}{t}\right)}\times log\ a\left(1-\frac{1}{t^2}\right)}{a\left(t+\frac{1}{t}\right)^{a-1}\left(1-\frac{1}{t^2}\right)}\\ \frac{dy}{dx}=\frac{a^{\left(t+\frac{1}{t}\right)}\times log\ a}{a\left(t+\frac{1}{t}\right)^{a-1}}
Summary
Exercise 11.7 Set 2 in RD Sharma's Class 12 Mathematics textbook continues to explore applications of derivatives in solving problems related to rates of change. This set introduces more complex scenarios and real-world applications, often involving multiple variables or intricate relationships. Students are challenged to apply their knowledge of differentiation techniques, including implicit differentiation and the chain rule, to analyze how various quantities change in relation to each other over time or with respect to other variables.
