第12类RD Sharma解决方案-第11章区分-练习11.4 |套装1

在以下每个中找到dy / dx：

问题1. xy = c ²

解决方案：

We have xy=c²

Differentiating both sides with respect to x.

d(xy)/dx = d(c²)/dx

By product rule,

y+x*dy/dx=0

Therefore the answer is.

 dy/dx=-y/x

问题2。y ^{3 -3xy 2} = x ³ + 3x ² y

解决方案：

We have 

y³-3xy²=x³+3x²y

Differentiating both sides with respect to x,

d(y³-3xy²)/dx=d(x³+3x²y)/dx

By product rule,

=> 3y²dy/dx-3y²-6xydy/dx=3x²+3x²dy/dx+6xy

=>3y²dy/dx-6xydy/dx-3x²dy/dx=3x²+3y²+6xy

=>dy/dx(3y²-3x²-6xy)=3x²+3y²+6xy

=>3dy/dx(y²-x²-2xy)=3(x²+y²+2xy)

=> dy/dx={3(x+y)²}/{3(y²-x²-2xy)

Therefore the answer is,

dy/dx=(x+y)²/(y²-x²-2xy)

问题3. x ^2/3 + y ^2/3 = a ^2/3

解决方案：

We have,

x^2/3+y^2/3=a^2/3

Differentiating both sides with respect to x,

d(x^2/3)/dx +d(y^2/3)/dx=d(a^2/3)/dx

=> 2/3x^1/3 +(2/3y^1/3)dy/dx=0

=>1/x^1/3 +(1/y^1/3)dy/dx =0

=> dy/dx=-y^1/3/x^1/3

Therefore the answer is,

dy/dx=-y^1/3/x^1/3

问题4. 4x + 3y = log(4x-3y)

解决方案：

We have,

4x+3y= log(4x-3y)

Differentiating both sides with respect to x,

d(4x+3y)/dx=d(log(4x-3y))/dx

=>4+3dy/dx=(1/(4x-3y))(4-3dy/dx)

=>3dy/dx+3dy/dx(1/4x-3y)=4/(4x-3y)-4

=>(3dy/dx)(1+1/(4x-3y))=(4-16x+12y)/(4x-3y)

=>(3dy/dx)((4x-3y+1)/(4x-3y))=(4-16x+12y)/(4x-3y)

=>(3dy/dx)(4x-3y+1)=4-16x+12y

=>(3dy/4dx)(4x-3y+1)=3y-4x+1

=>dy/dx=(4/3)((3y-4x+1)/(4x-3y+1))

Therefore the answer is,

dy/dx=4(3y-4x+1)/3(4x-3y+1)

问题5.(x ² / a ²⁾ +(y ² / b ² )= 1

解决方案：

We have,

(x²/a²)+(y²/b²)=1

Differentiating both sides with respect to x,

d(x²/a²)/dx +d(y²/b²)/dx =d(1)/dx

=>(2x/a²)+(2y/b²)(dy/dx)=0

=>(y/b²)(dy/dx)=-x/a²

=>dy/dx = -xb²/ya²

Therefore the answer is,

dy/dx =-xb²/ya².

问题6. x ⁵ + y ⁵ = 5xy

解决方案：

We have,

x⁵+y⁵ =5xy

Differentiating both sides with respect to x,

d(x⁵)/dx +d(y⁵)/dx=d(5xy)/ dx

=> 5x⁴ + 5y⁴dy/dx=5y+ (5x)dy/dx

=>y⁴⁽dy/dx)-x(dy/dx)=y-x⁴

=>dy/dx(y⁴-x)=y-x⁴

=>dy/dx=(y-x⁴)/(y⁴-x)

Therefore the answer is,

dy/dx=(y-x⁴)/(y⁴-x)

问题7.(x + y) ² = 2轴

解决方案：

We have,

(x+y)²=2axy

Differentiating with respect to x,

d(x+y)²/dx=d(2axy)/dx

=>2(x+y)(1+dy/dx)=2ax(dy/dx) +2ay

=>(x+y)+(x+y)dy/dx =ax(dy/dx)+ay

=>(x+y)dy/dx-ax(dy/dx)=ay-x-y()

=>(dy/dx)(x+y-ax)=ay-x-y

=>dy/dx=(ay-x-y)/(x+y-ax)

Therefore  the answer is,

dy/dx=(ay-x-y)/(x+y-ax)

问题8.(x ² + y ² ) ² = xy

解决方案：

We have,

(x²+y²)²=xy

Differentiating both sides with respect to x,

d(x²+y²)²/dx=d(xy)/dx

=>2(x²+y²)(2x+2y(dy/dx))=y+x(dy/dx)

=>4x(x²+y²)+4y(x²+y²)(dy/dx)=y+x(dy/dx)

=>4y(x²+y²)(dy/dx)-x(dy/dx)=y-4x(x²+y²)

=>(dy/dx)(4y(x²+y²)-x)=y-4x(x²+y²)

=>dy/dx=(y-4x(x²+y²))/(4y(x²+y²)-x)

Therefore the answer is,

dy/dx=(y-4x(x²+y²))/(4y(x²+y²)-x)

问题9. tan ^-1 (x ² + y ² )= a

解决方案：

We have,

tan^-1(x²+y²)=a

Differentiating both sides with respect to x ,

d(tan^-1(x²+y²))/dx=da/dx

=>(1/(x²+y²))(2x+2y(dy/dx))=0

=>x+y(dy/dx)=0

=> dy/dx=-x/y

Therefore the answer is,

dy/dx=-x/y

问题10. e ^xy = log(x / y)

解决方案：

We have,

e^x-y=log(x/y)

=>e^x-y=log x -log y

Differentiating both sides with respect to x,

d(e^x-y)/dx=d(log x- log y)/dx

=>e^x-y(1-dy/dx)=1/x-(1/y)(dy/dx)

=>e^x-y -e^x-y(dy/dx)=1/x -(1/y)(dy/dx)

=>(1/y)(dy/dx) – e^x-y(dy/dx)=1/x-e^x-y

=> dy/dx((1/y)-e^x-y)=(1-xe^x-y)/x

=> (dy/dx)(1-ye^x-y)/y=(1-xe^x-y)/x

=>dy/dx=y(1-xe^x-y)/x(1-ye^x-y)

Therefore the answer is,

dy/dx=y(1-xe^x-y)/x(1-ye^x-y)

问题11. sin(xy)+ cos(x + y)= 1

解决方案：

We have,

sin(xy)+ cos(x+y)=1

Differentiating both sides with respect to x,

d(sin(xy))/dx + d(cos(x+y))/dx=d1/dx

=>cos(xy)(y+xdy/dx) +(-sin(x+y)(1+dy/dx)= 0

=>cos(xy)(y+xdy/dx) = (sin(x+y)(1+dy/dx)

=>ycos(xy)+x*cos(xy)*(dy/dx)= sin(x+y) + sin(x+y)* (dy/dx)

=>x*cos(xy)*(dy/dx) – sin(x+y)* (dy/dx) = sin(x+y) – ycos(xy)

=>(dy/dx)((x*cos(xy))-sin(x+y))= sin(x+y) – ycos(xy)

=>dy/dx =(sin(x+y)-ycos(xy))/((x*cos(xy))-sin(x+y))

Therefore, the answer is,

dy/dx=(sin(x+y)-ycos(xy))/((x*cos(xy))-sin(x+y))

问题12.(1-x ² ) ^1/2 +(1-y ² ) ^1/2 = a(xy)

解决方案：

We have,

(1-x²)^1/2+(1-y²)^1/2=a(x-y)

Let x=sin A and y= sin B

So the expression becomes,

cosA + cosB=a(sinA-sinB)

=>a=(cosA+cosB)/(sinA-sinB)

=>a=(2(cos((A+B)/2))*(cos((A-B)/2)))/(2cos((A+B)/2)*sin((A-B)/2)))

=> a =(cos(A-B)/2)/(sin(A-B)/2)

=> a=cot((A-B)/2)

=>cot^-1a=((A-B)/2)

=>2cot^-1a=((A-B)/2)

Differentiating both sides with respect to x,

d(2cot^-1a)/dx=d(A-B)/dx

=>0=d(sin^-1x)/dx -d(sin^-1y)/dx

=> 0 = 1/((1-x²)^1/2) -(1/(1-y²)^1/2)*(dy/dx)

=>(1/(1-y²)^1/2)*dy/dx=1/((1-x²)^1/2)

=>dy/dx=((1-y²)^1/2)/(1-x²)^1/2

Therefore, the answer is,

dy/dx=((1-y²)^1/2)/(1-x²)^1/2

问题13. y(1-x ² ) ^1/2 + x(1-y ² ) ^1/2 = 1

解决方案：

We have,

y(1-x²)^1/2+x(1-y²)^1/2=1

Let, x=sin A and y=sin B

So, the expression becomes,

(sin B)*(cos A)+(sin A)*(cos B) =1

=> sin(A+B) =1

=> sin^-1(1) =A+B

=>A+B =22/(7*2)

=>sin^-1x +sin^-1y=22/14

Differentiating both sides with respect to x,

d(sin^-1x)/dx +d(sin^-1 y)/dx=d(22/14)/dx

=>1/((1-x²)^1/2)+ (1/((1-y²)^1/2))(dy/dx)=0

=>dy/dx=-((1-y²)^1/2)/((1-x²)^1/2)

Therefore, the answer is,

dy/dx=-((1-y²)^1/2)/((1-x²)^1/2)

问题14.如果xy = 1，则证明dy / dx + y ² = 0

解决方案：

We have,

xy=1

Differentiating both sides with respect to x,

d(xy)/dx =d1/dx

=>x(dy/dx)+y=0

=>dy/dx =-y/x

Also x=1/y

so,   dy/dx=-y(y)

=>dy/dx+y²=0

Hence, proved.

问题15.如果xy ² = 1，则证明2(dy / dx)+ y ³ = 0

解决方案：

We have,

xy²=1

Differentiating with respect to x,

d(xy²)/dx=d1/dx

=>2xy(dy/dx)+y² =0

=>dy/dx=-y²/2xy

=>dy/dx =-y/2x

Also x=1/y²

So, dy/dx=-y(y²)/2

=>2dy/dx=-y³

2dy/d+y³=0

Hence, proved.