Question

5. (3.5pts) Given the equation y = y²x+2 cos(x) find . Show each step of your work.
dx

Answer 1

Answer:

$\dfrac{dy}{dx}=\dfrac{y^2-2 \sin(x)}{1-2xy}$

Step-by-step explanation:

Given equation:

$y=y^2x+2 \cos(x)$

To find the derivative of the given equation, use implicit differentiation.

Add d/dx in front of each term:

$\dfrac{d}{dx}\:y =\dfrac{d}{dx}\:y^2x+\dfrac{d}{dx}\:2 \cos(x)$

Differentiate terms in x only (and constant terms) with respect to x:

$\dfrac{d}{dx}\:y =\dfrac{d}{dx}\:y^2x-2 \sin(x)$

Use the chain rule to differentiate terms in y only:

(in practice this means to differentiate with respect to y then place dy/dx on the end)

$\dfrac{dy}{dx}\: =\dfrac{d}{dx}\:y^2x-2 \sin(x)$

Use the product rule on the term in x and y:

$\textsf{let }u=y^2 \implies \dfrac{du}{dx}=2y \dfrac{dy}{dx}$

$\textsf{let }v=x \implies \dfrac{dv}{dx}=1$

$\implies u\dfrac{dv}{dx}+v\dfrac{du}{dx}=y^2+2xy \dfrac{dy}{dx}$

Therefore:

$\dfrac{dy}{dx} =y^2+2xy \dfrac{dy}{dx}-2 \sin(x)$

Rearrange to make dy/dx the subject:

$\dfrac{dy}{dx}-2xy \dfrac{dy}{dx} =y^2-2 \sin(x)$

$\dfrac{dy}{dx}(1-2xy)=y^2-2 \sin(x)$

$\dfrac{dy}{dx}=\dfrac{y^2-2 \sin(x)}{1-2xy}$