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implicit differentiation

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

Nov 20, 2018
edited by Guest  Nov 20, 2018

#1
+22307
+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:

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

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

Nov 20, 2018

#1
+22307
+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:

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

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

heureka Nov 20, 2018