Question #346256

Topic: Implicit Differentiation


1. Find y’ in 𝑥3 + 2𝑦3 = 3𝑥2𝑦.


2. Find the derivative of 𝑦 = √𝑠𝑖𝑛𝑥𝑦

1
Expert's answer
2022-05-30T23:29:38-0400

1.


x3+2y3=3x2yx^3 + 2y^3 =3x^2y

Differentiate both sides with respect to xx and us the Chain Rule


3x2+6y2y=6xy+3x2y3x^2+6y^2y'=6xy+3x^2y'

x2+2y2y=2xy+3x2yx^2+2y^2y'=2xy+3x^2y'

Solve for yy'


y=x22xy3x22y2y'=\dfrac{x^2-2xy}{3x^2-2y^2}


2.


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

Differentiate both sides with respect to xx and us the Chain Rule


y=12sin(xy)(sin(xy))y'=\dfrac{1}{2\sqrt{\sin(xy)}}(\sin(xy))'

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




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

2sin(xy)y=ycos(xy)+xcos(xy)y2\sqrt{\sin(xy)}y'=y\cos(xy)+x\cos(xy)y'

Solve for yy'

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




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