how do you simplify with expanding?

**delgeezee** · October 13th 2011, 12:47 PM

I am reading an example on mathlab and it is simplifying a derivative using the product rule and the chain rule. I just want to know how the example simplifies without expanding.

$\frac{dy}{dx}=(14(7x-2))(3-x^5)^2+(7x-2)^2(-10x^4(3-x^5))$

= $\frac{dy}{dx}=(2(7x-2)(3-x^5)[-5x^4(7x-2)+7(3-x^5)]$

**Soroban** · October 13th 2011, 01:28 PM

Hello, delgeezee!

We have: . $y \;=\;(7x-2)^2(3-x^5)^2$

Product/Chain Rule: . $\frac{dy}{dx} \;=\;2\!\cdot\!(7x-2)\!\cdot\!7\!\cdot\!(3-x^5)^2 + (7x-2)^2\!\cdot\!2(3-x^5)\!\cdot\!(-5x^4)$

Simplify a bit: . . . . . . $\frac{dy}{dx} \;=\;14(7x-2)(3-x^5)^2 - 10x^4(7x-2)^2(3-x^5)$

We have two groups: . $\frac{dy}{dx} \;=\;\underbrace{14(7x-2)(3-x^5)^2} - \underbrace{10x^4(7x-2)^2(3-x^5)}$

They have common factors: . $2,\;(7x-2),\;\text{and }(3-x^5)$

Factor them out!

. . $\frac{dy}{dx} \;=\;2(7x-2)(3-x^5)\,\bigg[7(3-x^5) - 5x^4(7x-2)\bigg]$

. m . . $=\;2(7x-2)(3-x^5)\,\big[21 - 7x^5 - 35x^5 + 10x^4\big]$

. m . . $=\;2(7x-2)(3-x^5)(21 + 10x^4 - 42x^5)$

**delgeezee** · October 13th 2011, 02:05 PM

Aha, now that makes sense!

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