Question #57107

2 Answers
Nov 28, 2017

dy/dx=-6cos^2(2x)sin(2x)+3cos(3x)

Explanation:

.

y=cos^3(2x)+sin(3x)

dy/dx=3cos^2(2x)(-sin(2x))(2)+cos(3x)(3)=

dy/dx=-6cos^2(2x)sin(2x)+3cos(3x)

Nov 29, 2017

dy/dx=-6cos^2(2x)sin(2x)+3cos(3x)

Explanation:

.

y=cos^3(2x)+sin(3x)

We have to take the derivetive of each term separately:

Let's do cos^3(2x) first:

Let's say y_1=cos^3(2x). Note that this is cos of another function which is 2x.

Let's put u=2x. Then du=2dx

(du)/dx=2

cosu itself is another function.

Let's say z=cosu. Now, let's take the derivative:

dz/(du)=-sinu

Also, plugging cosu into the original function give us:

y_1=z^3

dy_1/dz=3z^2

The Chain Rule says:

dy_1/dx=dy_1/dz*dz/(du)*(du)/dx Let's plug them in:

dy_1/dx=(3z^2)(-sinu)(2) Now, let's substitute back for z:

dy_1/dx=(3cos^2u)(-sinu)(2) Now, let's substitute back for u:

dy_1/dx=3cos^2(2x)(-sin(2x))(2) Now, we simplify:

dy_1/dx=-6cos^2(2x)sin(2x)

Now, we do the second term:

Let's say y_2=sin(3x) But 3x is another function.

Let's say w=3x Then (dw)/dx=3 and

y_2=sinw

dy_2/(dw)=cosw

The Chain Rule says:

dy_2/dx=dy_2/(dw)*(dw)/dx Now, let's plug them in:

dy_2/dx=(cosw)(3)=3cosw Now, let's substitute back for w:

dy_2/dx=3cos(3x)

According to our own definition:

y=y_1+y_2 Then:

dy/dx=dy_1/dx+dy_2/dx

Now. we can plug the derivatives of both terms into the above equation:

dy/dx=-6cos^2(2x)sin(2x)+3cos(3x)