Differentiating

[ArifRaihan]

Tried to differentiate f(x) = x^2(5x^2 - 2)^4
got 40x^3 (5x^2 - 2)^3 + 2x(5x^2 - 2)^4
but the final answer in the back of the book is 2x(5x^2 - 2)^3(25x^2-2)
and wolframalpha also got the same answer as the book

What is my mistake or the steps that i didnt complete yet?

 

[ArifRaihan]

Nevermind i got it now

 

[Dr.Peterson]

Nevermind i got it now

I presume you saw that your answer is equivalent to theirs; they just factored out 2x(5x^2 - 2)^3. Your answer would be acceptable, but what they did is a good habit.

 

[JeffM]

Just in case some student thinks that the OP’s method shows how to use substitution sensibly, it is not. The whole idea of substitution is to reduce things to simple discrete steps that reduce the risk of error.

[math]f(x) = x^2(5x^2 - 2)^4 = uv \text { where } u = x^2, \ v = w^4, \text { and } w = (5x^2 - 2) \implies\\ f’ = uv’ + u’v, \ u’ = 2x, \ v’ = 4w^3w’, \text { and } w‘ = 2 * 5x = 10x.\\ \therefore f’ = x^2 * 4w^3w’ + 2xv = x^2 * 4w^3 * 10x + 2xw^4 = \\ 2xw^3( 4 * 5 * x^2 + w) = 2xw^3(20x^2 + 5x^2 - 2) = 2x(x^2 - 2)^3(25x^2 - 2) [/math]
Notice how much simpler that is than the gyrations of the OP.

 

[lookagain]

Just in case some student thinks that the OP’s method shows how to use substitution sensibly, it is not. The whole idea of substitution is to reduce things to simple discrete steps that

> > > reduce the risk of error. < < <

[math]f(x) = x^2(5x^2 - 2)^4 = uv \text { where } u = x^2, \ v = w^4, \text { and } w = (5x^2 - 2) \implies\\ f’ = uv’ + u’v, \ u’ = 2x, \ v’ = 4w^3w’, \text { and } w‘ = 2 * 5x = 10x.\\ \therefore f’ = x^2 * 4w^3w’ + 2xv = x^2 * 4w^3 * 10x + 2xw^4 = \\ 2xw^3( 4 * 5 * x^2 + w) = 2xw^3(20x^2 + 5x^2 - 2) = 2x(x^2 - 2)^3(25x^2 - 2) [/math]
Notice how much simpler that is than the gyrations of the OP. Really!?

I don't find your way "much simpler" or even "simpler." You increased it by three
more variables. With more variable expressions, you are more likely to make an
error in copying. What's this!? You have a typo. The binomial that is cubed in the
answer should be \(\displaystyle \ 5x^2 - 2. \)

I'll give it to you that your solution should have less writing/fewer characters
when all is said and done, compared to the OP if he would take the top post and
keep working more steps until it is factored into the final book's answer.

However, I don't want to discourage you, JeffM, from using this method of yours.

 

[Dr.Peterson]

Just to be clear, nothing the OP did was wrong or extraneous:



They simply applied the product rule in a way that shows the details of that rule, as many students do when they first learn it. There is no substitution, just an identification of the "u" and "v" in the formula, and their derivatives.

 

[JeffM]

However, I don't want to discourage you, JeffM, from using this method of yours.

That’s so kind. I shall treasure it for life.