第12类NCERT解决方案-数学第I部分–第5章连续性和可微性–练习5.3

找$\frac{dy}{dx}$以下的：

问题1. 2x + 3y = sin x

解决方案：

On differentiating both sides w.r.t. x, we get

2 + 3 $\frac{dy}{dx}$ = cos x

3 $\frac{dy}{dx}$ = cos x – 2

$\frac{dy}{dx}$ = (cosx – 2)/3 

问题2。2x + 3y = sin y

解决方案：

On differentiating both sides w.r.t. x, we get

2 + 3 $\frac{dy}{dx}$ = cos y$\frac{dy}{dx}$

(cosy – 3) $\frac{dy}{dx}$ = 2

$\frac{dy}{dx} = \frac{2}{(cos⁡y - 3)}$

问题3. ax + ² = cos y

解决方案：

On differentiating both sides w.r.t. x, we get

a + b * 2y($\frac{dy}{dx}$) = -sin y * $\frac{dy}{dx}$

(2by + siny) $\frac{dy}{dx}$ = -a

$\frac{dy}{dx} = \frac{-a}{2by+siny}$

问题4. xy + y ² =棕褐色x + y

解决方案：

On differentiating both sides w.r.t. x, we get

(x *$\frac{dy}{dx}$ + y) + 2y$\frac{dy}{dx}$ = sec²x + $\frac{dy}{dx}$

(x + 2y – 1)$\frac{dy}{dx}$ = sec²x – y

$\frac{dy}{dx} = \frac{(sec^2x-y)}{(x+2y-1)}$

问题5. x ² + xy + y ² = 100

解决方案：

On differentiating both sides w.r.t. x, we get

2x + (x$\frac{dy}{dx}$ + y) + 2y$\frac{dy}{dx}$ = 0

(x + 2y) * $\frac{dy}{dx}$ = -(2x + y)

$\frac{dy}{dx} = \frac{-(2x + y)}{(x + 2y)}$ 

问题6. x ³ + x ² y + xy ² + y ³ = 81

解决方案：

Differentiate both sides w.r.t. x

3x²+(x²$\frac{dy}{dx}$ + y * 2x) + (x * 2y * $\frac{dy}{dx}$ + y²) + 3y² * $\frac{dy}{dx}$ = 0

(x² + 2xy + 3y²)$\frac{dy}{dx}$ = -(3x² + 2xy + y²)

$\frac{dy}{dx} = \frac{3x^2 + 2xy + y^2}{x^2 + 2xy + 3y^2}$

问题7：Sin ² y + cos xy =π

解决方案：

Differentiate both sides w.r.t. x

2 sin y * $\frac{dy}{dx}$ (sin⁡y) – sin(xy) * $\frac{dy}{dx}$ xy = 0

2sin y * cosy $\frac{dy}{dx}$ – sin(xy)(x * $\frac{dy}{dx}$ + y) = 0

(2sin cos y – sin (xy) – x)) $\frac{dy}{dx}$ = y(xy)

$\frac{dy}{dx} = \frac{y sin⁡(xy)}{2 sin⁡ y cos⁡ y - x sin⁡ (xy)}$

$\frac{dy}{dx} = \frac{ysin(xy)}{sin2y - x sinxy}$

问题8：sin ² x + cos ² y = 1

解决方案：

 2 sin x * $\frac{dy}{dx}$ (sin ⁡x) + 2 cos y * $\frac{dy}{dx}$ (cos ⁡y) = 0

2 sin x * cos x + 2 cos y*(-sin y) * $\frac{dy}{dx}$ = 0

2 sin x * cos x – 2 cos x – 2 cos y sin y * $\frac{dy}{dx}$ = 0

Sin(2x) – sin(2y) – $\frac{dy}{dx}$ = 0

$\frac{dy}{dx} = \frac{sin(2x)}{sin(2y)}$

问题9：y = sin ^-1 (\ frac {2x} {(1 + x ² )}

解决方案：

Put x = tanθ          

θ = tan^-1x

$y = sin^{-1}\frac{2tanθ}{1+tan^2θ}$

y = sin^-1(sin 2θ)          

y = 2θ

y = 2tan^-1x         -(1)

On differentiating eq(1), we get

$\frac{dy}{dx} = \frac{2}{(1 + x^2)}$

问题10。 $Y=tan^{-1}\frac{(3x-x^3)}{(1-3x^2 )}$ ，-1 /√3

解决方案：

Put x = tanθ         

θ = tan^-1x

y = $tan^{-1}(\frac{3tanθ - tan^3θ}{1-3tan^2θ})$

y = tan^-1(tan 3θ)         

y = 3θ

y = 3tan^-1x          -(1)

On differentiating eq(1), we get

$\frac{dy}{dx} = \frac{3}{(1 + x^2)}$

问题11 $y = cos^{-1}(\frac{1-x^2}{1+x^2})$ ，0

解决方案：

Put x = tanθ          

θ  = tan^-1 x

y = $cos^{-1}(\frac{1-tan^2θ}{1+tan^2θ})$

y = cos^-1(cos 2θ)   

y = 2θ 

y = 2tan^-1x         -(1)

On differentiating eq(1), we get

$\frac{dy}{dx} = \frac{2}{(1 + x^2)}$  

问题12。 $y = sin^{-1}(\frac{1-x^2}{1+x^2})$ ，0

解决方案：

Put x = tanθ       

θ = tan^-1x

$y = sin^{-1}(\frac{1 - tan^2θ}{1 + tan^2θ})$

y = sin^-1(cos 2θ)

y = sin^-1(sin (π/2 – 2θ))         

y = π/2 – 2θ

y = π/2 – 2 tan^-1x  

$\frac{dy}{dx} = \frac{(-2)}{(1 + x^2)}$

问题13。 $y = cos^{-1}(\frac{2x}{1+x^2})$ ，-1

解决方案：

Put x = tanθ     

θ = tan^-1x

y = cos^{-1}($\frac{2tanθ}{1 + tan^2θ}$)        

y = cos^-1(sin 2θ)

y = cos^-1(cos (π/2 – 2θ))         

y = π/2 – 2θ

y = π/2 – 2tan^-1x  

$\frac{dy}{dx} = \frac{(-2)}{(1 + x^2)}$

问题14。 $y = sin^{-1}(2x\sqrt{1-x^2})$ ，-1 /√2

解决方案：

Put x = sinθ     

θ =  sin^-1 x

y = sin^-1(2sinθ√(1 – sin²θ))

y = sin^-1(sin 2θ) = 2θ

y = 2sin^-1x

$\frac{dy}{dx} = \frac{2}{\sqrt{(1 - x^2)}}$

问题15。 $y = sec^{-1}(\frac{1}{2x^2-1})$ ，0

解决方案：

Put x = tanθ

$y = sec^{-1}(\frac{1}{2 cos^2 - 1})$       

y = sec^-1(1/cos2θ))

y = sec^-1(sec2θ) = 2θ

y = 2cos^-1x

$\frac{dy}{dx}$ = $\frac{-2}{\sqrt{1 - x^2} }$