We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
184
1
avatar

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

 Nov 20, 2018
edited by Guest  Nov 20, 2018

Best Answer 

 #1
avatar+23516 
+12

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

 Nov 20, 2018
 #1
avatar+23516 
+12
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

5 Online Users

avatar