How do you use implicit differentiation to find #(dy)/(dx)# given #sin2x^2y^3=3x^3+1#?

Question

Luke Alvarez · Accepted Answer

# dy/dx = (9x^2 - 4xcos(2x^2)y^3)/(3sin(2x^2)y^2) #

When we differentiate #y# wrt #x# we get #dy/dx#. However, we only differentiate explicit functions of #y# wrt #x#. But if we apply the chain rule we can differentiate an implicit function of #y# wrt #y# but we must also multiply the result by #dy/dx#. Example: #d/dx(y^2) = d/dy(y^2)dy/dx = 2ydy/dx # When this is done in situ it is known as implicit differentiation. Now, we have: # sin2x^2y^3=3x^3+1 # Implicitly differentiating wrt #x# (applying product rule): # (sin2x^2)(d/dx y^3) + (d/dx sin2x^2)(y^3) = 9x^2# # :. (sin2x^2)(3y^2dy/dx ) + (4xcos2x^2)(y^3) = 9x^2# # :. 3sin(2x^2)y^2dy/dx + 4xcos(2x^2)y^3 = 9x^2# # :. 3sin(2x^2)y^2dy/dx = 9x^2 - 4xcos(2x^2)y^3 # # :. dy/dx = (9x^2 - 4xcos(2x^2)y^3)/(3sin(2x^2)y^2) # Advanced Calculus There is another (often faster) approach using partial derivatives. Suppose we cannot find #y# explicitly as a function of #x#, only implicitly through the equation #F(x, y) = 0# which defines #y# as a function of #x, y = y(x)#. Therefore we can write #F(x, y) = 0# as #F(x, y(x)) = 0#. Differentiating both sides of this, using the partial chain rule gives us # (partial F)/(partial x) (1) + (partial F)/(partial y) dy/dx = 0 => dy/dx = −((partial F)/(partial x)) / ((partial F)/(partial y)) # So Let # F(x,y) = sin2x^2y^3-3x^3-1 #; Then; #(partial F)/(partial x) = 4xcos(2x^2)y^3-9x^2 # #(partial F)/(partial y) = sin(2x^2)(3y^2) # And so: # dy/dx = -(4xcos(2x^2)y^3-9x^2)/(sin(2x^2)(3y^2)) # # " " = (9x^2-4xcos(2x^2)y^3)/(3sin(2x^2)y^2) #, as before.

Isabella Ackerman · Answer

undefined To find $(dy)/(dx)$ using implicit differentiation, follow these steps:

1. Differentiate both sides of the equation with respect to $x$.
2. Apply the chain rule whenever you differentiate terms involving $y$.
3. Solve the resulting equation for $(dy)/(dx)$.

Applying these steps to the given equation $sin(2x^2y^3) = 3x^3 + 1$, we get:

$$
\frac{d}{dx}(sin(2x^2y^3)) = \frac{d}{dx}(3x^3 + 1)
$$

This simplifies to:

$$
cos(2x^2y^3) \cdot \frac{d}{dx}(2x^2y^3) = 9x^2
$$

Then, using the chain rule:

$$
cos(2x^2y^3) \cdot (4x^3y^2 \cdot \frac{dy}{dx} + 6x^2y^2) = 9x^2
$$

Solve for $(dy)/(dx)$:

$$
\frac{dy}{dx} = \frac{9x^2 - 6x^2y^2 \cdot cos(2x^2y^3)}{4x^3y^2 \cdot cos(2x^2y^3)}
$$

This is the derivative of $y$ with respect to $x$.