How do you find the Implicit differentiation of #x^4-5xy^3+y^6=21#?

Question

Isabella Ackerman · Accepted Answer

You use the fact that #y# is a function of #x#;

Samantha Adkison · Answer

Tedious, but worth it: If #x# and #y# are related by: #x^4-5xy^3+y^6=21#, then we can find #y'# (or find #dy/dx)# in either of two ways. We understand that #y# is some function (or functions) of #x#.  I mean, if we choose an #x# value, we could find (or at least try to find) the value of #y# that makes the equation true. Our choices are 1) make the funcion(s) explicit (obvious), or 2) leave the function(s) implcit (hidden) and find #y'# without first solving for #y#. Gio has shown us how to fin #y'# without first solving for #y#, and that answers the question about implicit differentiation.  I want to show the explicit function(s). We'll solve:
#x^4-5xy^3+y^6=21#, for #y#. #x^4-5xy^3+y^6=21#, #y^6-5xy^3+x^4-21=0#, This equation is quadratic in #y^3#, with #a=1#, #b=5x#, and #c=x^4-21#. Using the quadratic formula, we get: #y^3= (-(-5x) +- sqrt((-5x)^2-4(1)(x^4-21)))/(2(1))# #y^3= (5x +- sqrt(25x^2-4x^4+84))/2# So  #y=root(3)((5x +- sqrt(25x^2-4x^4+84))/2)# Now, if we wanted to differentiate, we would get the equivalent of  #y' = dy/dx = (5y^3-4x^3)/(6y^5-15xy^2)#

Christopher Agresta · Answer

undefined To find the implicit differentiation of $x^4 - 5xy^3 + y^6 = 21$, follow these steps:

1. Differentiate each term of the equation with respect to $x$.
2. Apply the chain rule when differentiating terms involving $y$ with respect to $x$.
3. Solve for $\frac{dy}{dx}$.

The implicit differentiation of $x^4 - 5xy^3 + y^6 = 21$ yields:

$$4x^3 - 5y^3 - 15xy^2 \frac{dy}{dx} + 6y^5 \frac{dy}{dx} = 0$$

Rearranging terms, we get:

$$4x^3 - 15xy^2 \frac{dy}{dx} + 6y^5 \frac{dy}{dx} = 5y^3$$

Combine like terms:

$$4x^3 + (6y^5 - 15xy^2) \frac{dy}{dx} = 5y^3$$

Finally, solve for $\frac{dy}{dx}$:

$$\frac{dy}{dx} = \frac{5y^3 - 4x^3}{6y^5 - 15xy^2}$$

So, the implicit differentiation of $x^4 - 5xy^3 + y^6 = 21$ is $\frac{dy}{dx} = \frac{5y^3 - 4x^3}{6y^5 - 15xy^2}$.