Calculus Question

Question

Find the derivative of y = cos³sin²2x.

 

Help?!

Answer 1

Start by applying the product rule. If y= fg then y' = f'g + fg'

Let f = cos³x and g = sin²2x

 

Figure out the derivatives of each individual function f and g

 

The derivative of f also uses the product rule

(d/dx) (cosx cosx cosx) = -sinx cosx cosx -sinx cosx cosx -sinx cosx cosx

f' = -3cos²sinx 

 

The derivative of g uses the chain rule. Know (d/dx)sin²x = sin(2x)

Then (d/dx) sin²(2x) = 2sin(4x)

 

Now, put it all together. y' = f'g + fg'

 

y' =(-3cos²xsinx)(sin²(2x)) + (cos³x)(2sin(4x))

 

 

Answer 2

If you meant

 

y = (cos³x)sin²2x

 

then use the product rule, power rule, and chain rule.

 

dy/dx = (3cos²x)(-sinx)sin²2x + (cos³x)[2sin(2x)][cos(2x)(2)]

 

You can simplify it from here.

Answer 3

y = cos³sin²2x

 

There are a couple of things going on in this trig function; so, I am going to rewrite it so that you can see the applications of the chain rule and the power easier:

 

y= [ cos ( sin(2x) )² ]³

 

recall the chain rule:

if f(x) = g[ h(x) ] then f'(x) = g' [ h(x) ] · h'(x)

 

recall the power rule:

if f(x) = [ g(x) ]ⁿ then f'(x) = n[ g(x) ]^n-1 · g'(x)

 

 

y' = 3 [ cos ( sin(2x) )² ]² · -sin ( sin(2x) )² · 2 ( sin(2x) )¹ ·cos(2x) · 2

y' = -12 cos²(sin(2x) ) · sin²(sin(2x) ) · sin(2x) · cos(2x)