If #f(x)= sin3x # and #g(x) = 2x^2 -3x #, how do you differentiate #f(g(x)) # using the chain rule?

Question

If #f(x)=  sin3x # and #g(x)  =  2x^2  -3x  #, how do you differentiate #f(g(x))  # using the chain rule?

Scarlett Allison · Accepted Answer

#f'(g(x))*g'(x)=cos(6x^2-9x)*(12x-9)# We have #f(g(x))=sin(3(2x^2-3x))=sin(6x^2-9x)# By the chain rule #f'(g(x))*g'(x)# we get #cos(6x^2-9x)*(12x-9)#

Evelyn Agard · Answer

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

1. Find $ f'(g(x)) $, the derivative of $ f(x) $ with respect to $ g(x) $.
2. Find $ g'(x) $, the derivative of $ g(x) $ with respect to $ x $.
3. Multiply $ f'(g(x)) $ by $ g'(x) $ to get the final derivative.

Here's how to apply this to the given functions:

1. $ f(x) = \sin(3x) $
   $ f'(x) = 3\cos(3x) $ (derivative of $\sin(ax)$ is $a\cos(ax)$)

2. $ g(x) = 2x^2 - 3x $
   $ g'(x) = 4x - 3 $ (derivative of $2x^2$ is $4x$, derivative of $-3x$ is $-3$)

3. Now, substitute $ g(x) $ into $ f(x) $ to get $ f(g(x)) $:
   $ f(g(x)) = \sin(3(2x^2 - 3x)) = \sin(6x^2 - 9x) $

4. Apply the chain rule:
   $ f'(g(x)) = 3\cos(6x^2 - 9x) $

5. Multiply by $ g'(x) $:
   $ f'(g(x)) \cdot g'(x) = (3\cos(6x^2 - 9x)) \cdot (4x - 3) $

So, the derivative of $ f(g(x)) $ with respect to $ x $ using the chain rule is $ (3\cos(6x^2 - 9x)) \cdot (4x - 3) $.