How do you differentiate #f(x)=cos(-3x^2+2)^2#?

Question

HIX Tutor · Accepted Answer

undefined To differentiate $ f(x) = \cos(-3x^2 + 2)^2 $, follow these steps:

1. **Identify the outer and inner functions.**  
   Here, the outer function is $ u^2 $ where $ u = \cos(-3x^2 + 2) $.

2. **Differentiate the outer function.**  
   The derivative of $ u^2 $ is $ 2u \cdot u' $.

3. **Find $ u' $.**  
   Differentiate $ u = \cos(-3x^2 + 2) $.  
   The derivative of $ \cos(v) $ is $ -\sin(v) \cdot v' $.  
   Here, $ v = -3x^2 + 2 $, so $ v' = -6x $.

4. **Put it together.**  
   Now we have:  
   $$ u' = -\sin(-3x^2 + 2) \cdot (-6x) = 6x \sin(-3x^2 + 2) $$

5. **Combine the results.**  
   Using the chain rule:  
   $$ f'(x) = 2u \cdot u' $$  
   $$ f'(x) = 2\cos(-3x^2 + 2) \cdot 6x \sin(-3x^2 + 2) $$

6. **Final answer.**  
   $$ f'(x) = 12x \cos(-3x^2 + 2) \sin(-3x^2 + 2) $$

That's it! You have found the derivative.

Christopher Allers · Answer

#f'(x)=12x(-3x^2+2) sin(-3x^2+2)^2# #f(x)=cos(-3x^2+2)^2# #f'(x)=-sin(-3x^2+2)^2 * 2(-3x^2+2) *-6x# #f'(x)=12x(-3x^2+2) sin(-3x^2+2)^2#

Aurora Alvarado · Answer

undefined To differentiate $ f(x) = \cos(-3x^2 + 2)^2 $, apply the chain rule and power rule:

$ f'(x) = -2\cos(-3x^2 + 2)\sin(-3x^2 + 2)(-6x) $