#### To determine

**To establish:** The rule dc=0 for working with differential, where *c* is a constant.

#### Explanation

Consider the function y=c.

The differential is dy=f′(x)dx.

The derivative of the function f(x)=c is f′(x)=0.

dy=(0)dxdy=0

That is, dc=0.

Hence, the required result is proved.

#### To determine

**To establish:** The rules d(cu)=c du for working with differential, where *c* is a constant and *u* is a function of *x.*

#### Explanation

Consider the function y=cu.

The differential is dy=f′(x)dx.

The derivative of the function f(x)=cu is, f′(x)=cdudx.

Substitute f′(x)=cdudx in dy=f′(x)dx,

dy=cdudxdxdy=c⋅du

That is, d(cu)=cdu.

Hence, the require result proved.

#### To determine

**To establish:** The rules d(u+v)= du+dv for working with differential.

#### Explanation

Consider the function y=u+v.

The differential is dy=f′(x)dx.

The derivative of the function f(x)=u+v is computed as follows,

f′(x)=ddx(u+v)=dudx+dvdx

Thus, the derivative of the function f(x)=u+v is f′(x)=dudx+dvdx.

Substitute f′(x)=dudx+dvdx in dy=f′(x)dx,

dy=(dudx+dvdx)dx=dudxdx+dvdxdx=du+dv

That is, d(u+v)=du+dv.

Hence, the require result proved.

#### To determine

**To establish:** The rules d(uv)=udv+ vdu for working with differential.

#### Explanation

Consider the function y=uv.

The differential is dy=f′(x)dx.

The derivative of the function f(x)=uv is computed as follows,

f′(x)=ddx(uv)=udvdx+vdudx

Thus, the derivative of the function f(x)=uv is f′(x)=udvdx+vdudx.

Substitute f′(x)=udvdx+vdudx in dy=f′(x)dx,

dy=(udvdx+vdudx)dx=udvdxdx+vdudxdx=udv+vdu

That is, d(uv)=udv+vdu.

Hence, the required result proved.

#### To determine

**To establish:** The rules d(uv)= vdu−udvv2 for working with differential.

#### Explanation

Consider the function y=uv.

The differential is dy=f′(x)dx.

The derivative of the function f(x)=uv is computed as follows,

f′(x)=ddx(uv)=vdudx−udvdxv2

Thus, the derivative of the function f(x)=uv is f′(x)=vdudx−udvdxv2.

Substitute f′(x)=vdudx−udvdxv2 in dy=f′(x)dx,

dy=(vdudx−udvdxv2)dx=(vdudx−udvdx)dxv2=vdudxdx−udvdxdxv2=vdu−udvv2

That is, d(uv)=vdu−udvv2.

Hence, the required result proved.

#### To determine

**To establish:** The rules d(xn)=nxn−1dx for working with differential.

#### Explanation

Consider the function y=xn.

The differential is dy=f′(x)dx.

The derivative of the function f(x)=xn is computed as follows,

f′(x)=ddx(xn)=nxn−1

Thus, the derivative of the function f(x)=xn is f′(x)=nxn−1.

Substitute f′(x)=nxn−1 in dy=f′(x)dx,

dy=(nxn−1)dx=nxn−1dx

That is, d(xn)=nxn−1dx.

Hence, the required result proved.