Evaluate and simplify y'.
xy⁴+x⁴y=1

Guida verificata passo dopo passo

First, recognize that the given equation is implicit: \( y \cdot x y^4 + x^4 y = 1 \). We need to differentiate both sides with respect to \( x \).

Apply the product rule to the term \( y \cdot x y^4 \). The product rule states that \( (uv)' = u'v + uv' \). Here, let \( u = y \) and \( v = x y^4 \). Differentiate \( u \) and \( v \) with respect to \( x \).

Differentiate \( u = y \) with respect to \( x \) to get \( u' = y' \). For \( v = x y^4 \), apply the product rule again: \( v' = (x)' y^4 + x (y^4)' \). Differentiate \( x \) to get 1, and \( y^4 \) using the chain rule to get \( 4y^3 y' \).

Now, differentiate the second term \( x^4 y \) using the product rule: \( (x^4 y)' = (x^4)' y + x^4 (y)' \). Differentiate \( x^4 \) to get \( 4x^3 \), and \( y \) to get \( y' \).

Combine all differentiated terms and set the derivative of the right side of the equation, which is 0, equal to the derivative of the left side. Solve for \( y' \) to find the expression for the derivative.

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Implicit Differentiation

Implicit differentiation is a technique used to differentiate equations where the dependent and independent variables are not explicitly separated. In this case, we differentiate both sides of the equation xy⁴ + x⁴y = 1 with respect to x, treating y as a function of x. This allows us to find the derivative y' without needing to solve for y explicitly.

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Percorso guidato

05:14

Finding The Implicit Derivative

Product Rule

The product rule is a fundamental rule in calculus used to differentiate products of two functions. It states that if you have two functions u(x) and v(x), the derivative of their product is given by u'v + uv'. In the context of the given equation, we will apply the product rule to differentiate terms like xy⁴ and x⁴y, where both x and y are functions of x.

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05:18

The Product Rule

Chain Rule

The chain rule is a method for differentiating composite functions. It states that if a variable y depends on u, which in turn depends on x, then the derivative of y with respect to x is the product of the derivative of y with respect to u and the derivative of u with respect to x. In this problem, we will use the chain rule to differentiate terms involving y, as y is implicitly defined in terms of x.

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