How do you implicitly differentiate #2=e^ysinx-x^2y^3 #?

Question

How do you implicitly differentiate #2=e^ysinx-x^2y^3  #?

Emily Ammerman · Accepted Answer

I found: #(dy)/(dx)=(2xy^3-e^ycos(x))/(e^ysin(x)-3x^(2)y^2)# You need to consider #y# as a function of #x# and so differentiate it accordingly. For example: if you have #y^2# differentiating you'll get #2y(dy)/(dx)# to consider the fact that #y# represents a function of something! in our case: #0=e^y(dy)/(dx)sin(x)+e^ycos(x)-2xy^3-x^(2)3y^2(dy)/(dx)# Here I used the Product Rule and every time I differentiated #y# I needed to include #(dy)/(dx)#: collect #(dy)/(dx)#: #(dy)/(dx)[e^ysin(x)-x^(2)3y^2]=2xy^3-e^ycos(x)# and: #(dy)/(dx)=(2xy^3-e^ycos(x))/(e^ysin(x)-x^(2)3y^2)#

Cameron Alexander · Answer

undefined To implicitly differentiate $2 = e^{y} \sin{x} - x^{2} y^{3}$, follow these steps:

1. Differentiate both sides of the equation with respect to $x$.
2. Apply the chain rule when differentiating $e^{y}$.
3. Apply the product rule when differentiating $x^{2} y^{3}$.
4. Solve for $\frac{dy}{dx}$ after differentiation.

Here's the step-by-step process:

1. $0 = e^{y} \cos{x} \frac{dy}{dx} - 2x y^{3} - 3x^{2} y^{2} \frac{dy}{dx}$
2. $0 = e^{y} \cos{x} \frac{dy}{dx} - 3x^{2} y^{2} \frac{dy}{dx} - 2x y^{3}$
3. Factor out $\frac{dy}{dx}$:  
   $0 = (\cos{x} e^{y} - 3x^{2} y^{2}) \frac{dy}{dx} - 2x y^{3}$
4. Solve for $\frac{dy}{dx}$:  
   $\frac{dy}{dx} = \frac{2x y^{3}}{\cos{x} e^{y} - 3x^{2} y^{2}}$