Find dy/dx of (2x-1)^3 * (x+7)^-3

[warwick]

I just had an exam. I used the Chain Rule twice but did not get one of the choices listed in the multiple choice problem. Obviously, I did something wrong.

f(x) = (2x-1)^3 * (x+7)^-3

f'(x) = (2x-1)^3 [-3(x+7)^-4 (1)] + (x+7)^-3 [3(2x-1)^2 (2)]

 

[galactus]

You probably had trouble with the simplification.

Product rule: \(\displaystyle \L\\(2x-1)^{3}(-3)(x+7)^{-4}+(x+7)^{-3}(3)(2x-1)^{2}(2)\)

\(\displaystyle \L\\\frac{-3(2x-1)^{3}}{(x+7)^{4}}+\frac{6(2x-1)^{2}}{(x+7)^{3}}\)

Multiply top and bottom of the right half by (x+7) to make the denominators equal.

\(\displaystyle \L\\\frac{-3(2x-1)^{3}}{(x+7)^{4}}+\frac{6(2x-1)^{2}}{(x+7)^{3}}\cdot\frac{(x+7)}{(x+7)}\)

This simplifies to:

\(\displaystyle \H\\\frac{45(2x-1)^{2}}{(x+7)^{4}}\)

Is that one of your choices?.

 

[warwick]

I actually wrote it with the (x-7) terms in the denominator. I'm trying to figure out the simplification right now.

 

[galactus]

It's just a matter of practice and seeing what to do. When you have two denominators like these, look and ask yourself what you need to do to make them equal. In this case we multiplied one side by (x+7).

For the numerator, you have a common factor of \(\displaystyle (2x-1)^{2}\). Factor it out.

\(\displaystyle (2x-1)^{2}(6x+42-3(2x-1))\)

\(\displaystyle 45(2x-1)^{2}\)

 

[soroban]

Hello, warwick!

I just had an exam. I used the Chain Rule twice
but did not get one of the choices listed in the multiple choice problem.
Obviously, I did something wrong. . . . Yes

\(\displaystyle f(x) \:= \2x\,-\,1)^3(x\,+\,7)^{-3}\)

\(\displaystyle f'(x) \:= \2x\,-\,1)^3(-3)(x\,+\,7)^{-4} (1) \,+ \,(x\,+\,7)^{-3}(3)(2x\,-\,1)^2 (2)\)

You didn't even try to simplify, did you?

We have: \(\displaystyle \,f'(x)\:=\:-3(2x\,-\,1)^3(x\,+\,7)^{-4}\,+\,6(2x\,-\,1)^2(x\,+\,7)^{-3}\)

If that was one of the answer-choices, you wouldn't have recognized it
. . because you didn't know enough to multiply \(\displaystyle 3\,\times\,2\)

They usually simplify further . . .

Factor: \(\displaystyle \:3(2x\,-\,1)^2(x\,+\,7)^{-4}\,\left[-(2x\,-\,1)\,+\,2(x\,+\,7)\right]\)

. . . .\(\displaystyle = \;3(2x\,-\,1)^2(x\,+\,y)^{-4}\,\left[-2x\,+\,1\,+\,2x\,+\,14\right]\)

. . . .\(\displaystyle = \;3(2x\,-\,1)^2(x\,+\,7)^{-4}\,(15)\)

. . . .\(\displaystyle = \;45(2x\,-\,1)^2(x\,+\,y)^{-4}\) . . . or: \(\displaystyle \L\,\frac{45(2x\,-\,1)^2}{(x\,+\,7)^4}\)

 

[warwick]

Ugh. I could've done this. That's what pisses me off. I knew I needed to factor and simply but just didn't follow it through.