# Math Help - Checking work (derivatives)

1. ## Checking work (derivatives)

I have the binomial

$(t^2+6)(1+t^2)$

working out the problem I have ...

$t^4+7t^2+6$

y'= $y'=4t^3+14t$

= $2t(2t^2+7)$

Is this correct?

Using the product rule, I get the same thing

$(2t)(1+t^2)+(2t)(t^2+6)$

= $2t+2t^3+2t^3+12t$ = $4t^3+14t$

= $2t(2t^2+7)$

Sorry, I got a multiple choice assignment and I want to make extra sure as the answers are all very close or changing a sign here or something subtle

2. Perfect!

3. Hello, XIII13Thirteen!

I have the function: . $y \;=\;(t^2+6)(1+t^2)$

Working out the problem I have:

. . $y \;=\;t^4+7t^2+6\quad\Rightarrow\quad y'\:=\:4t^3+14t \:=\:2t(2t^2+7)$

Is this correct? . . . . . Yes!

Using the product rule, I get the same thing:

$y' \;=\;(2t)(1+t^2)+(2t)(t^2+6) \:=\:2t+2t^3+2t^3+12t$

. . $= \:4t^3+14t \:=\:2t(2t^2+7)$ . . . . . (Yes!)²

That is an excellent way to check your work!

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I have told my students that they can easily create practice problems
. . for the Product Rule and Quotient Rule.

We know that: . $y \;=\;x^9\quad\Rightarrow\quad y' \;=\;9x^8$

$\text{Now use the Product Rule on: }\;y \;=\;\overbrace{(x^4)}^{f(x)}\overbrace{(x^5)}^{g( x)}$

$\text{We have: }\;y'\;=\;\overbrace{(x^4)}^{f(x)}\overbrace{(5x^4 )}^{g'(x)} + \overbrace{(4x^3)}^{f'(x)}\overbrace{(x^5)}^{g(x)} \;=\;5x^8 + 4x^8 \;=\;9x^8\qquad\hdots\;see?$

$\text{Now use the Quotient Rule on: }\;y \;=\;\frac{\overbrace{x^{12}}^{f(x)}}{\underbrace{ x^3}_{g(x)}}$

$\text{We have: }\;y' \;=\;\frac{\overbrace{(x^3)}^{g(x)}\overbrace{(12x ^{11}}^{f'(x)}) - \overbrace{(x^{12})}^{f(x)}\overbrace{(3x^2)}^{g'( x)}}{\underbrace{(x^3)^2}_{[g(x)]^2}} \;=\;\frac{12x^{14}-3x^{14}}{x^6} \;=\;\frac{9x^{14}}{x^6} \;=\;9x^8$

4. @ Soroban: Thanks for giving me an extended view of the concepts as well. I think the more I drill this stuff into myself the more confidant I'll feel about it