Answer to Question #346256 in Calculus for rei

Question

Topic: Implicit Differentiation

1. Find y’ in 𝑥3 + 2𝑦3 = 3𝑥2𝑦.

2. Find the derivative of 𝑦 = √𝑠𝑖𝑛𝑥𝑦

Answer 1

1.

$$x^3 + 2y^3 =3x^2y$$

Differentiate both sides with respect to $x$ and us the Chain Rule

$$3x^2+6y^2y'=6xy+3x^2y'$$

$$x^2+2y^2y'=2xy+3x^2y'$$

Solve for $y'$

$$y'=\dfrac{x^2-2xy}{3x^2-2y^2}$$

2.

$$y =\sqrt{\sin(xy)}$$

Differentiate both sides with respect to $x$ and us the Chain Rule

$$y'=\dfrac{1}{2\sqrt{\sin(xy)}}(\sin(xy))'$$

$$y'=\dfrac{\cos(xy)}{2\sqrt{\sin(xy)}}(xy)'$$

$$y'=\dfrac{\cos(xy)(y+xy')}{2\sqrt{\sin(xy)}}$$

$$2\sqrt{\sin(xy)}y'=y\cos(xy)+x\cos(xy)y'$$

Solve for $y'$

$$y'=\dfrac{y\cos(xy)}{2\sqrt{\sin(xy)}-x\cos(xy)}$$