This is a very long solution process, so bare with me:
#A#: We are trying to find the derivative of the product of two functions, so it helps to use the Product Rule
#f'(x)g(x)+f(x)g'(x)#
where #f(x)=sin(3x)# and #g(x)=cos(5x)#
By use of the Chain Rule, we know that
#f'(x)=3cos(3x)# and #g'(x)=-5sin(5x)#
Plugging into our expression for the Product Rule, we get
#(3cos3x)(cos5x)+(sin3x)(-5sin5x)#
Which can be rewritten as
#B#: #3cos(3x)cos(5x)-5sin(3x)sin(5x)#
Finding the Second Derivative:
If we define the following:
#color(red)(a=3cos(3x)cos(5x))#
#color(blue)(b=5sin(3x)sin(5x))#
Then we essentially have
#C#: #d/dx (a-b)=d/dx(a)-d/dx(b)#
We can find the derivative of #a#, then the derivative of #b#, and then subtract them.
#D#: Finding the derivative of #color(red)(a=(3cos(3x)cos(5x)))#
We have a product of functions again, so we can use the Product Rule. We have
#f(x)=3cos(3x)# and #g(x)=cos(5x)#
From another application of the Chain Rule, we know
#f'(x)=-9sin(3x)# and #g'(x)=-5sin(5x)#
Plugging into our expression for the Product Rule, we get
#color(red)(-9sin(3x)cos(5x)-3cos(3x)5sin(5x))#
Recall that this business is only the derivative of #a#. Now, we must find the derivative of #b#.
#E#: Finding the Derivative of #color(blue)(b=5sin(3x)sin(5x))#
We know
#f(x)=5sin(3x)=>f'(x)=15cos(3x)# (Chain Rule)
#g(x)=sin(5x)=>g'(x)=5cos(5x)# (Chain Rule)
Recall that the Product Rule is
#f'(x)g(x)+f(x)g'(x)#
Plugging our values in, we get
#color(blue)(15cos(3x)sin(5x)+5sin(3x)5cos(5x))#
The derivative of the expression in step B (second derivative of original expression) is
#color(red)((-9sin(3x)cos(5x)-3cos(3x)5sin(5x)))-color(blue)((15cos(3x)sin(5x)+5sin(3x)5cos(5x)))#
#F#: Simplifying the expression
#bar(ul|color(white)(2/2)-34sin(3x)cos(5x)-30cos(3x)sin(5x)color(white)(2/2)|)#
This was an extremely hairy problem, but to reiterate, we
#A#. Used Product/Chain Rule to find First Derivative
#B#. Defined variables to simplify our differentiation
#C#. Rewrote the derivative of the difference as the difference of the derivatives
#D+E#. Found derivatives separately to find Second Derivative (with Product/Chain Rules)
#F#. Simplified the result
I really hope I didn't make a careless mistake, but if I did, one of our Socratic community members may rectify it.
Hope this helps!
