How do you differentiate #f(x) =((65e^-7x)+2)^3 # using the chain rule?

Question

1525 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-03T09:31:35

#823875e^98x^2+50700e^49x+780e^-7#, or to make it simpler:

#195e^-7((65e^-7x)+2)^2#

We have to find:

#d/dx((65e^-7x)+2)^3#

We can say that this is in the form #u^3#, where #u=(65e^-7x)+2#.

To solve this, then, we must compute #d/(du)u^3# and multiply that by #d/dx((65e^-7x)+2)#

As #d/dxx^n=nx^(n-1)#, we have:

#3u^2*d/dx((65e^-7x)+2)#

#3u^2*d/dx(65e^-7x)#

#3u^2*65e^-7*d/dxx#

#3u^2*65e^-7#

And as #u=(65e^-7x)+2#, we have:

#195e^-7((65e^-7x)+2)^2#

What I do after this is just expansion. It isn't needed, and it just makes it more confusing.

#195e^-7((65e^-7x)^2+2(2)(65e^-7x)+2^2)#

#195e^-7(4225e^-14x^2+260e^-7x+4)#

#823875e^98x^2+50700e^49x+780e^-7#