+0  
 
0
30
1
avatar

if xy3=8, find dy/dx?????

Guest Nov 20, 2018
edited by Guest  Nov 20, 2018

Best Answer 

 #1
avatar+20578 
+9

if xy3=8, find dy/dx ?

 

1. Solution:

Differentiate both sides of the equation, getting : \(\begin{array}{|lll|} \hline \text{Differentiate both sides of the equation, getting: } \\ \qquad D(xy^3) &= D(8) \\ \text{Use the product rule}: \\ \qquad \{ xD(y^3)+D(x)y^3\} &= 0 \quad & | \quad D(8) = 0 \\ \qquad \{ x(3y^2y') + (1)\cdot y^3 \} &= 0 \quad & | \quad : y^2 \\ \qquad 3xy' + y &= 0 \quad & | \quad : -y \\ \qquad 3xy' &= -y \quad & | \quad : 3x \\ \qquad \mathbf{ y' } & \mathbf{=-\dfrac{ y } {3x} } \\ \hline \end{array} \)

 

2. Solution:

\(\text{General formula for derivative of implicit function} \\ \text{ If ${\displaystyle R(x,y)=0,}$ the derivative of the implicit function $y(x)$ is given by } \\ \text{$\dfrac{dy}{dx} = -\dfrac{R_x}{R_y}$} \\ \text{with $R_x = \dfrac{\partial R}{\partial x}$ and $R_y = \dfrac{\partial R}{\partial y}$ } \)

\(\begin{array}{|lrcll|} \hline \text{Begin with: } \\ & xy^3 -8 &=& 0 \\ & R_x &=& 1\cdot y^3 \\ & R_y &=& x\cdot 3y^2 \\ & \dfrac{dy}{dx} &=& -\dfrac{R_x}{R_y} \\ & &=& -\dfrac{y^3}{x\cdot 3y^2} \\ & \mathbf{ \dfrac{dy}{dx} } & \mathbf{=} & \mathbf{-\dfrac{y}{3x}} \\ \hline \end{array}\)

 

laugh

heureka  Nov 20, 2018
 #1
avatar+20578 
+9
Best Answer

if xy3=8, find dy/dx ?

 

1. Solution:

Differentiate both sides of the equation, getting : \(\begin{array}{|lll|} \hline \text{Differentiate both sides of the equation, getting: } \\ \qquad D(xy^3) &= D(8) \\ \text{Use the product rule}: \\ \qquad \{ xD(y^3)+D(x)y^3\} &= 0 \quad & | \quad D(8) = 0 \\ \qquad \{ x(3y^2y') + (1)\cdot y^3 \} &= 0 \quad & | \quad : y^2 \\ \qquad 3xy' + y &= 0 \quad & | \quad : -y \\ \qquad 3xy' &= -y \quad & | \quad : 3x \\ \qquad \mathbf{ y' } & \mathbf{=-\dfrac{ y } {3x} } \\ \hline \end{array} \)

 

2. Solution:

\(\text{General formula for derivative of implicit function} \\ \text{ If ${\displaystyle R(x,y)=0,}$ the derivative of the implicit function $y(x)$ is given by } \\ \text{$\dfrac{dy}{dx} = -\dfrac{R_x}{R_y}$} \\ \text{with $R_x = \dfrac{\partial R}{\partial x}$ and $R_y = \dfrac{\partial R}{\partial y}$ } \)

\(\begin{array}{|lrcll|} \hline \text{Begin with: } \\ & xy^3 -8 &=& 0 \\ & R_x &=& 1\cdot y^3 \\ & R_y &=& x\cdot 3y^2 \\ & \dfrac{dy}{dx} &=& -\dfrac{R_x}{R_y} \\ & &=& -\dfrac{y^3}{x\cdot 3y^2} \\ & \mathbf{ \dfrac{dy}{dx} } & \mathbf{=} & \mathbf{-\dfrac{y}{3x}} \\ \hline \end{array}\)

 

laugh

heureka  Nov 20, 2018

10 Online Users

avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.