第 12 类 RD Sharma 解决方案 - 第 11 章微分 - 练习 11.7 |设置 2

问题 11. 查找$\frac{dy}{dx}$ ， 什么时候$x=\frac{2t}{1+t^2}$和$y=\frac{1-t^2}{1+t^2}$

解决方案：

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)

$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-4t}{(1+t^2)^2}\times\frac{(1+t^2)^2}{2(1-t^2)}\\ =\frac{-2t}{1-t^2}\\ \frac{dy}{dx}=-\frac{x}{y}\ \ \ \ \left[Since,\ \frac{x}{y}=\frac{2t}{1+t^2}\times\frac{1+t^2}{1-t^2}=\frac{2t}{1-t^2}\right]$

问题 12. 查找$\frac{dy}{dx}$ ， 什么时候$x=cos^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right)$和$y=sin^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right)$

解决方案：

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)

$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{1}{(1+t^2)}\times\frac{(1+t^2)}{-1}\\ \frac{dy}{dx}=1$

问题 13. 查找$\frac{dy}{dx}$ ， 什么时候$x=\frac{1-t^2}{1+t^2}$和$y=\frac{2t}{1+t^2}$

解决方案：

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,

$\frac{dy}{dt}=\left[\frac{(1+t^2)\frac{d}{dt}(2t)-2t\frac{d}{dt}(1+t^2)}{(1+t^2)^2}\right]\\ =\left[\frac{(1+t^2)(2)-2t(2t)}{(1+t^2)^2}\right]\\ =\left[\frac{2+2t^2-4t^2}{(1+t^2)^2}\right]\\ \frac{dx}{dt}=\frac{2(1-t^2)}{(1+t^2)^2}\ \ \ \ \ ....(2)$

问题 14. 如果 x = 2co sθ – cos2θ 和 y = 2sinθ – sin2θ ，证明$\frac{dy}{dx}=tan\left(\frac{3\theta}{2}\right)$

解决方案：

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),

$\frac{\frac{dy}{dθ}}{\frac{dx}{dθ}}=\frac{2(cosθ -cos2θ )}{2(sin2θ - sinθ )}\\ =\frac{cosθ -cos2θ}{sin2θ -sinθ}\\ \frac{dy}{dx}=\frac{-2sin\left(\frac{θ +2θ}{2}\right)sin\left(\frac{θ -2θ}{2}\right)}{2cos\left(\frac{2θ +θ}{2}\right)sin\left(\frac{2θ -θ}{2}\right)}\ \ \ \ \left[Since,\ sinA-sinB=2cos\left(\frac{A+B}{2}\right)sin\left(\frac{A-B}{2}\right),\ \ cosA-cosB=-2sin\left(\frac{A+B}{2}\right)sin\left(\frac{A-B}{2}\right)\right]\\ =\frac{-sin\left(\frac{3θ )}{2}\right)\left(sin\left(\frac{-θ}{2}\right)\right)}{cos\left(\frac{3θ}{2}\right)sin\left(\frac{θ}{2}\right)}\\ =\frac{sin\left(\frac{3θ}{2}\right)}{cos\left(\frac{3θ}{2}\right)}\\ \frac{dy}{dx}=tan\left(\frac{3θ}{2}\right)$

问题 15. 如果 x = e ^cos2t和 y = e ^sin2t证明， $\frac{dy}{dx}=\frac{y\ log\ x}{x\ log\ y}$

解决方案：

Here,

x = e^cos2t

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 = e^sin2t

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)

$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2cos2te^{sin2t}}{-2sin2te^{cos2t}}\\ \frac{dy}{dx}=-\frac{ylogx}{xlogy}$

$\left[Since,\ x=e^{cos2t}\Rightarrow logx=cos2t\ \ y=e^{sin2t}\Rightarrow logy=sin2t\right]$

问题 16. 如果 x = cos t 和 y = sin t，证明$\frac{dy}{dx}=\frac{1}{\sqrt3}\ at\ t=\frac{2x}{3}$

解决方案：

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),

$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{cos\ t}{-sin\ t}\\ \frac{dy}{dx}=-cot\ t\\ \left(\frac{dy}{dx}\right)=-cot\left(\frac{2\pi}{3}\right)\\ =-cot\left(\pi-\frac{\pi}{3}\right)\\ =-\left[-cot\left(\frac{\pi}{3}\right)\right]\\ =cot\left(\frac{\pi}{3}\right)\\ \frac{dy}{dx}=\frac{1}{\sqrt3}$

问题 17. 如果$x=a\left(t+\frac{1}{t}\right)$和$y=a\left(t-\frac{1}{t}\right)$ ， 证明$\frac{dy}{dx}=\frac{x}{y}$

解决方案：

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)

$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=a\frac{(t^2+1)}{t^2}\times\frac{t^2}{a(t^2-1)}\\ \frac{dy}{dx}=\frac{t^2+1}{t^2-1}\\ \frac{dy}{dx}=\frac{x}{y}\ \ \ \ \ \ \left[Since,\ \frac{x}{y}=\frac{a(t^2+1)}{t}\times\frac{t}{a(t^2-1)}=\left(\frac{t^2+1}{t^2-1}\right)\right]$

问题 18. 如果$x=sin^{-1}\left(\frac{2t}{1+t^2}\right)$和$y =tan^{-1}\left(\frac{2t}{1-t^2}\right)$ , -1 < 1 < 1，证明$\frac{dy}{dx}=1$

解决方案：

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),

$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2}{1+t^2}\times\frac{1+t^2}{2}\\ \frac{dy}{dx}=1$

问题 19. 如果 x 和 y 通过方程参数连接，不消去参数，求$\frac{dy}{dx}$ ， 什么时候： $x=\frac{sin^3t}{\sqrt{cos2t}}$ , $y=\frac{cos^3t}{\sqrt{cos2t}}$

解决方案：

Here, the given equations are$x=\frac{sin^3t}{\sqrt{cos2t}}$and$y=\frac{cos^3t}{\sqrt{cos2t}}$

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,

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-3cos\ 2t.cos^2t. sin\ t+cos^3t\ sin\ 2t}{3cos\ 2t\ sin^2t\ cos\ t+sin^3t\ sin\ 2t}\\ =\frac{-3cos\ 2t\ cos^2t.sin\ t+cos^3t(2sin\ t\ cos\ t)}{3cos\ 2t\ sin^2t\ cos\ t+sin^3t(2sin\ t+2sin^3t)}\\ =\frac{[-3(2cos^2t-1)cos\ t+2cos^3t]}{[3(1-2sin^2t)sin\ t+2sin^3t]}\ \ \ \ \ \ \ [cos\ 2t=(2cos^2t-1),\ cos\ 2t=(1-2sin^2t)]\\ =\frac{-4cos^3t+3cos\ t}{3sin\ t-4sin^3\ t}\\ =\frac{-cos\ 3t}{sin\ 3t}\ \ \ \ \ \ [cos\ 3t=4cos^3t3cos\ t,\ sin\ 3t=3sin\ t-4sin^3t]\\ =-cot3t$

问题 20. 如果$x=\left(t+\frac{1}{t}\right)^a$和$y=a^{\left(t+\frac{1}{t}\right)}$ ， 找$\frac{dy}{dx}$

解决方案：

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)

$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{a^{\left(t+\frac{1}{t}\right)}\times log\ a\left(1-\frac{1}{t^2}\right)}{a\left(t+\frac{1}{t}\right)^{a-1}\left(1-\frac{1}{t^2}\right)}\\ \frac{dy}{dx}=\frac{a^{\left(t+\frac{1}{t}\right)}\times log\ a}{a\left(t+\frac{1}{t}\right)^{a-1}}$