Mathematics: Frame a quadratic equation when 2 zeros are 1/3and -1/2

Frame a quadratic equation when 2 zeros are 1/3and

Ответ:

cjjohnson1221

15.06.2020

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Ответ:

ZachLaVine2016

23.06.2021

Mathematics

I'm thinking you want something that looks like this
$\frac{dy}{dx}\sqrt{x+\frac{1}{4}sin^2(2x)}$ or it could be like this
$\frac{dy}{dx}\sqrt{x}+\frac{1}{4}sin^2(2x)$

use chain rule
dy/dx f(g(x))=f'(g(x))g'(x)
and √x=x^(1/2)
so
I'll do the first one first
$\frac{dy}{dx}\sqrt{x+\frac{1}{4}sin^2(2x)}$ =
$\frac{dy}{dx}(x+\frac{1}{4}sin^2(2x))^{\frac{1}{2}}$ =
$\frac{1}{2(x+\frac{1}{4}sin^2(2x))^{\frac{1}{2}}}(1+sin(2x)cos(2x))$ =
$\frac{1+sin(2x)cos(2x)}{2\sqrt{x+\frac{1}{4}sin^2(2x}}$

the 2nd one is a bit simpler
$\frac{dy}{dx}\sqrt{x}+\frac{1}{4}sin^2(2x)$ =
$\frac{dy}{dx}x^{\frac{1}{2}}+\frac{1}{4}sin^2(2x)$ =
$\frac{1}{2}x^{\frac{-1}{2}}+\frac{1}{4}sin^2(2x)$ =
$\frac{1}{2x^{\frac{1}{2}}}+sin(2x)cos(2x)$ =
$\frac{1}{2\sqrt{x}}+sin(2x)cos(2x)$

in conclusion
$\frac{dy}{dx}\sqrt{x+\frac{1}{4}sin^2(2x)}$ = $\frac{1+sin(2x)cos(2x)}{2\sqrt{x+\frac{1}{4}sin^2(2x}}$ and
$\frac{dy}{dx}\sqrt{x}+\frac{1}{4}sin^2(2x)$ = $\frac{1}{2\sqrt{x}}+sin(2x)cos(2x)$
depends which one it was

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Frame a quadratic equation when 2 zeros are 1/3and -1/2

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