If #f(x)= 2 x^2 + x # and #g(x) = 2e^x + 1 #, how do you differentiate #f(g(x)) # using the chain rule?

Question

If #f(x)=  2  x^2  +   x # and #g(x)  = 2e^x +  1    #, how do you differentiate #f(g(x))  # using the chain rule?

Isaiah Allen · Accepted Answer

#16e^(2x)+10e^x# #f(g(x)# #=f(2e^x+1)=2(2e^x+1)^2+2e^x+1# #color(white)(xxxxxxxxx)=2(4e^(2x)+4e^x+1)+2e^x+1# #color(white)(xxxxxxxxx)=8e^(2x)+8e^x+2+2e^x+1# #color(white)(xxxxxxxxx)=8e^(2x)+10e^x+3# #rArrd/dx(8e^(2x)+10e^x+3)# #=8e^(2x)xxd/dx(2x)+10e^xlarrcolor(blue)"chain rule"# #=16e^(2x)+10e^x#

Isabella Ackerman · Answer

See below.

#f(x)=2x^2+x# , #g(x)=2e^x+1# #:.# #f(g(x))=2(2e^x+1)^2+2e^x+1=8e^(2x)+10e^x+3# Using the chain rule: #dy/dx=(dy)/(du)*(du)/dx# Let #u=e^x# #8u^2+10u+3# #dy/(dx)(8u^2)=dy/(du)(8u^2) * (du)/(dx)(u)=16u*e^x=16e^(2x)# #dy/(dx)(10u)=dy/(du)(10u) * (du)/(dx)(u)=10 * e^x=10e^x# #:.# #dy/dx(8x^(2x)+10e^x+3)=16e^(2x)+10e^x#

Isabella Ackerman · Answer

undefined To differentiate $ f(g(x)) $ using the chain rule, first find $ f'(x) $ and $ g'(x) $, then substitute $ g(x) $ into $ f'(x) $ and multiply by $ g'(x) $.

Given $ f(x) = 2x^2 + x $ and $ g(x) = 2e^x + 1 $:

1. Find $ f'(x) $:
   $ f'(x) = 4x + 1 $

2. Find $ g'(x) $:
   $ g'(x) = 2e^x $

3. Substitute $ g(x) $ into $ f'(x) $:
   $ f'(g(x)) = 4g(x) + 1 $

4. Multiply by $ g'(x) $:
   $ f'(g(x)) \cdot g'(x) = (4g(x) + 1) \cdot 2e^x $

Therefore, the derivative of $ f(g(x)) $ using the chain rule is $ (4g(x) + 1) \cdot 2e^x $.