If #f(x)= cos 4 x # and #g(x) = -3x #, how do you differentiate #f(g(x)) # using the chain rule?

Question

If #f(x)=  cos 4  x # and #g(x)  =    -3x  #, how do you differentiate #f(g(x))  # using the chain rule?

Elijah Abram · Accepted Answer

#d/dx(f(g(x))=-12sin12x#

As #f(x)=cos4x# and #g(x)=-3x#, #f(g(x))=cos4(-3x)#

According to chain rule

#d/dxf(g(x))=(df)/(dg)xx(dg)/(dx)#

Hence, as #f(g(x))=cos4(-3x)#

#d/dx(f(g(x))=-4sin4(-3x) xx(-3)#

= #12sin(-12x)=-12sin12x#

Isaiah Allen · Answer

#-12sin(12x)# #(df)/(dx)=(df)/(dg)(dg)/(dx)# #=-4sin(4g(x))xx(-3)# #=12sin(4(-3x))# #-12sin(12x)#

Evelyn Agard · Answer

undefined To differentiate $ f(g(x)) $ using the chain rule, follow these steps:

1. Identify the outer function $ f(x) $ and the inner function $ g(x) $.
2. Calculate the derivative of the outer function $ f'(x) $.
3. Calculate the derivative of the inner function $ g'(x) $.
4. Substitute $ g(x) $ and $ g'(x) $ into $ f'(g(x)) $.
5. Multiply $ f'(g(x)) $ by $ g'(x) $ to get the final result.

Given $ f(x) = \cos^4(x) $ and $ g(x) = -3x $:

1. The outer function is $ f(x) = \cos^4(x) $ and the inner function is $ g(x) = -3x $.
2. The derivative of the outer function is $ f'(x) = 4\cos^3(x) \cdot (-\sin(x)) $.
3. The derivative of the inner function is $ g'(x) = -3 $.
4. Substitute $ g(x) = -3x $ into $ f'(x) $ to get $ f'(-3x) = 4\cos^3(-3x) \cdot (-\sin(-3x)) $.
5. Multiply $ f'(-3x) $ by $ g'(x) = -3 $ to get the final result:

$$ f'(g(x)) \cdot g'(x) = 4\cos^3(-3x) \cdot (-\sin(-3x)) \cdot (-3) $$

Therefore, the derivative of $ f(g(x)) $ with respect to $ x $ is:

$$ -12\cos^3(-3x) \cdot \sin(-3x) $$