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Prove the product rule for derivatives of complex valued function.

Short Answer

Expert verified

The solution is ddt(z1z2)=z2ddt(z1)+z1ddt(z2).

Step by step solution

01

Simplify the complex function z1(t)z2(t). 

Consider the complex valued functionsz1(t)andz2(t)wherez1(t)=x1(t)+iy1(t)andz2(t)=x2(t)+iy2(t).

Multiply the equations z1(t)=x1(t)+iy1(t)andz2(t)=x2(t)+iy2(t) as follows.

z1(t)z2(t)={x1(t)+iy1(t)}{x2(t)+iy2(t)}z1(t)z2(t)=x1(t)x2(t)−y1(t)y2(t)+i{x1(t)y2(t)+x2(t)y1(t)}

02

Determine the value of z2(t)dz1(t)dt+z1(t)dz2(t)dt . 

Differentiate the equationz1(t)=x1(t)+iy1(t) both side with respect to tas follows.

z1(t)=x1(t)+iy1(t)dz1(t)dt=dx1(t)dt+idy1(t)dt

Differentiate the equation z2(t)=x2(t)+iy2(t)both side with respect t to as follows.

z2(t)=x2(t)+iy2(t)dz2(t)dt=dx2(t)dt+idy2(t)dt

Substitute the values x1(t)+iy1(t)forz1(t) ,dx1(t)dt+idy1(t)dt for dz1(t)dt, x2(t)+iy2(t)forz2(t) anddx2(t)dt+idy2(t)dt fordz2(t)dt in z2(t)ddt(z1(t))+z1(t)ddt(z2(t))as follows.

z2(t)ddt(z1(t))+z1(t)ddt(z2(t))=[{x2(t)+iy2(t)}{dx1(t)dt+idy1(t)dt}+{x1(t)+iy1(t)}{dx2(t)dt+idy2(t)dt}]z2(t)ddt(z1(t))+z1(t)ddt(z2(t))=[{x2(t)dx1(t)dt+ix2(t)dy1(t)dt+iy2(t)dx1(t)dt−y2(t)dy1(t)dt}+{x1(t)dx2(t)dt+ix1(t)dy2(t)dt+iy1(t)dx2(t)dt−y1(t)dy2(t)dt}]z2(t)ddt(z1(t))+z1(t)ddt(z2(t))=[{x2(t)dx1(t)dt−y2(t)dy1(t)dt+i{x2(t)dy1(t)dt+y2(t)dx1(t)dt}}+{x1(t)dx2(t)dt−y1(t)dy2(t)dt+i{x1(t)dy2(t)dt+y1(t)dx2(t)dt}}]z2(t)ddt(z1(t))+z1(t)ddt(z2(t))=[{x2(t)dx1(t)dt−y2(t)dy1(t)dtx1(t)dx2(t)dt−y1(t)dy2(t)dt}+i{x2(t)dy1(t)dt+y2(t)dx1(t)dt+x1(t)dy2(t)dt+y1(t)dx2(t)dt}]

Therefore, the value of z2(t)ddt(z1(t))+z1(t)ddt(z2(t))is [{x2(t)dx1(t)dt−y2(t)dy1(t)dtx1(t)dx2(t)dt−y1(t)dy2(t)dt}+i{x2(t)dy1(t)dt+y2(t)dx1(t)dt+x1(t)dy2(t)dt+y1(t)dx2(t)dt}].

03

Determine the value of  ddt{z1(t)z2(t)}

Differentiate the equationz1(t)z2(t)=x1(t)x2(t)−y1(t)y2(t)+i{x1(t)y2(t)+x2(t)y1(t)} both side with respect to tas follows.

z1(t)z2(t)=x1(t)x2(t)−y1(t)y2(t)+i{x1(t)y2(t)+x2(t)y1(t)}ddt{z1(t)z2(t)}=ddt{x1(t)x2(t)−y1(t)y2(t)+i{x1(t)y2(t)+x2(t)y1(t)}}ddt{z1(t)z2(t)}=[x1(t)ddtx2(t)+x2(t)ddtx1(t)−y1(t)ddty2(t)−y2(t)ddty1(t)+i{x1(t)ddty2(t)+y2(t)ddtx1(t)+y1(t)ddtx2(t)+x2(t)ddty1(t)}]

04

Show that ddt{z1(t)z2(t)}=z2(t)ddt{z1(t)}+z1(t)ddt{z2(t)}.

As ddt{z1(t)z2(t)}=[x1(t)ddtx2(t)+x2(t)ddtx1(t)−y1(t)ddty2(t)−y2(t)ddty1(t)+i{x1(t)ddty2(t)+y2(t)ddtx1(t)+y1(t)ddtx2(t)+x2(t)ddty1(t)}], substitute the values z2(t)ddt(z1(t))+z1(t)ddt(z2(t))for [x1(t)ddtx2(t)+x2(t)ddtx1(t)−y1(t)ddty2(t)−y2(t)ddty1(t)+i{x1(t)ddty2(t)+y2(t)ddtx1(t)+y1(t)ddtx2(t)+x2(t)ddty1(t)}]in the equation ddt{z1(t)z2(t)}=[x1(t)ddtx2(t)+x2(t)ddtx1(t)−y1(t)ddty2(t)−y2(t)ddty1(t)+i{x1(t)ddty2(t)+y2(t)ddtx1(t)+y1(t)ddtx2(t)+x2(t)ddty1(t)}]as follows.

ddt{z1(t)z2(t)}=[x1(t)ddtx2(t)+x2(t)ddtx1(t)−y1(t)ddty2(t)−y2(t)ddty1(t)+i{x1(t)ddty2(t)+y2(t)ddtx1(t)+y1(t)ddtx2(t)+x2(t)ddty1(t)}]=z2(t)ddt(z1(t))+z1(t)ddt(z2(t))ddt{z1(t)z2(t)}=z2(t)ddt(z1(t))+z1(t)ddt(z2(t))

Hence, the product rule for derivative formula for the complex valued function is derived

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Most popular questions from this chapter

Consider a wooden block in the shape of a cube whose edges are 10 cm long. The density of the wood is 0.8 g /cm2 . The block is submersed in water; a guiding mechanism guarantees that the top and the bottom surfaces of the block are parallel to the surface of the water at all times. Let x(t)be the depth of the block in the water at time t. Assume that xis between 0 and 10 at all times.

a.Two forces are acting on the block: its weight and the buoyancy (the weight of the displaced water).

Recall that the density of water is 1 g/cm 3. Find formulas for these two forces.

b.Set up a differential equation for x(t). Find the solution, assuming that the block is initially completely submersed [x(0)=10] and at rest.

c.How does the period of the oscillation change if you change the dimensions of the block? (Consider a larger or smaller cube.) What if the wood has a different density or if the initial state is different? What if you conduct the experiment on the moon?

Solve the differential equation dxdt+3x=7and find the solution of the differential equation.

Find the real solution of the system dx→dt=[0-330]x→

Let be an matrix anda scalar. Consider the following two systems:

dx→dt=Ax→(l)dc→dt=(A+kln)c→(ll)

Show that if x→(t) is a solution of the system (l)then role="math" localid="1659701582223" c→(t)=ektx→(t)is a solution of the system (ll).

Consider a linear system dx→dt=Ax→of arbitrary size. Suppose x1(t)and x2(t)are a solution of the system. Is the sumx→(t)=x→1(t)+x→2(t) a solution as well? How do you know?

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