Finding the dirivative

[nikita.s]

"Find the dirivative, and give your answer in simplified form."

y=x^4(2x-5)^6

I was given a small handfull of these questions to do, but I can't do any of them. I don't even know how to start correctly. I tried finding g(x), g'(x), h(x), h'(x) then applying the product rule, but I got the wrong answer. Can someone give me a nudge in the right direction?

 

[soroban]

Hello, nikita.s!

It's possible that you had the right answer,
\(\displaystyle \;\;\)but their answer may have been simplified beyond all recognition.

Find the derivative, and give your answer in simplified form.
\(\displaystyle \;\;\;y \:=\:x^4(2x\,-\,5)^6\)

Product Rule: \(\displaystyle \,y' \;= \;\underbrac{x^4}\underbrace{\cdot6(2x\,-\,5)^5\cdot 2}\,+\,\underbrace{4x^3}\cdot\underbrace{(2x\,-\,5)^6}\)
. . . . . . . . . . . . . . . . .\(\displaystyle g(x)\;\;\;h'(x)\;\;\;\;\;g'(x)\;\;\,h(x)\)

. . . . . . . . . . . \(\displaystyle y' \;=\;12x^4(2x\,-\,5)^5\,+\,4x^3(2x\,-\,5)^6\)

Factor: . . . . . \(\displaystyle y' \;= \;4x^3(2x\,-\,5)^5\,\left[3x\,+\,(2x\,-\,5)\right]\)

We have: . . .\(\displaystyle y' \;= \;4x^3(2x\,-\,5)^5(5x\,-\,5)\)

Factor: . . . . . \(\displaystyle y' \;= \;4x^3(2x\,-\,5)^5\cdot5(x\,-\,1)\)

Therefore: . . \(\displaystyle y' \;= \;20x^3(2x\,-\,5)^5\)
.

 

[nikita.s]

I understand it a little better, thank you very much! Though I'm having trouble understanding how to factor that huge equation (first one), and also, from the second factoring step to the last, where did the (x-1) go? It looks like the 4x^3 and the 5 were multiplied together but the (x-1) just disappeared.

 

[ChaoticLlama]

I understand it a little better, thank you very much! Though I'm having trouble understanding how to factor that huge equation (first one), and also, from the second factoring step to the last, where did the (x-1) go? It looks like the 4x^3 and the 5 were multiplied together but the (x-1) just disappeared.

you are quite correct. Soroban merely forgot the (x-1) in the last line.

the answer should read as: y' = 20x³(x-1)(2x-5)^5

 

[nikita.s]

Oh, alright. Thank you!