Question

Implicit differentiation with rational exponents Determine the slope of the following curves at the given point. $$\sqrt[3]{x}+\sqrt[3]{y^{4}}=2 ;(1,1)$$

Short answer

Answer: The slope of the tangent line at the point (1, 1) is $$-\frac{1}{4}$$.

Step-by-step solution

Write down the given equation

We have the equation $$\sqrt[3]{x}+\sqrt[3]{y^{4}}=2$$ at the point \((1,1)\).

Differentiate implicitly with respect to x

To differentiate implicitly, we apply the chain rule to both terms on the left side of the equation with respect to x. For the first term, we get: $$\frac{d}{dx}(\sqrt[3]{x})=\frac{1}{3}x^{-\frac{2}{3}}$$ For the second term, we get: $$\frac{d}{dx}(\sqrt[3]{y^{4}})=\frac{d}{dx}(y^{4/3})=\frac{4}{3}y^{\frac{1}{3}}\cdot\frac{dy}{dx}$$ Now we can rewrite the given equation after differentiating both terms: $$\frac{1}{3}x^{-\frac{2}{3}}+\frac{4}{3}y^{\frac{1}{3}}\cdot\frac{dy}{dx}=0$$

Solve for dy/dx

Now we need to find the value of \(\frac{dy}{dx}\) at the given point \((1,1)\): $$\frac{dy}{dx} = -\frac{(\frac{1}{3})x^{-\frac{2}{3}}}{\frac{4}{3}y^{\frac{1}{3}}}$$ Substitute the coordinates \((1,1)\) into the equation: $$\frac{dy}{dx}\Big|_{(1,1)} = -\frac{(\frac{1}{3})(1)^{-\frac{2}{3}}}{\frac{4}{3}(1)^{\frac{1}{3}}}$$ Simplify: $$\frac{dy}{dx}\Big|_{(1,1)} = -\frac{\frac{1}{3}}{\frac{4}{3}}=-\frac{1}{4}$$

State the slope of the tangent line at the given point

The slope of the tangent line to the curve $$\sqrt[3]{x}+\sqrt[3]{y^{4}}=2$$ at the point \((1,1)\) is \(-\frac{1}{4}\).

Key concepts

Chain Rule

The chain rule is a fundamental concept in calculus. It allows us to differentiate composite functions; that is, functions within other functions.
For example, in the function \( f(g(x)) \), where \( f \) and \( g \) are separate functions, the chain rule gives us a way to find the derivative of this composite function.
We differentiate \( f(g(x)) \) by multiplying the derivative of the outer function \( f \) evaluated at \( g(x) \), by the derivative of the inner function \( g \) with respect to \( x \).

• Formula: If \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).
• This method is useful for finding derivatives of implicitly defined functions, like in our exercise.

Differentiating \( \sqrt[3]{y^4} \) using the chain rule, first think of it as \( (y^4)^{1/3} \).
Here, the outer function is \((u)^{1/3}\) and the inner function is \(u = y^4\). The derivative is then \( \frac{1}{3}(y^4)^{-2/3} \cdot 4y^3 \cdot \frac{dy}{dx}\).
Understanding the chain rule is essential when working on equations where one variable is defined implicitly in terms of another.

Rational Exponents

Rational exponents allow us to simplify expressions involving roots. A rational exponent is an exponent that is a fraction, where the numerator represents the power to which the base is raised, and the denominator represents the root.
For example, \( a^{m/n} = \sqrt[n]{a^m} \).

• Example: \( x^{1/3} = \sqrt[3]{x} \).
• Used to express roots as powers for easier differentiation and algebraic manipulation.

In the implicit differentiation problem, we deal with both \( \sqrt[3]{x} \) and \( \sqrt[3]{y^4} \) which translate into the terms \( x^{1/3} \) and \( (y^4)^{1/3} \) respectively.
Rewriting roots in this manner helps streamline the differentiation process by allowing us to apply standard rules for derivatives, such as the power rule.
When differentiating expressions with rational exponents, always apply the chain rule if the base is something other than a simple variable.

Tangent Line Slope

The slope of a tangent line to a curve at a particular point is crucial in calculus. It represents the instantaneous rate of change of the function at that point.
This concept is analogous to finding the slope of a straight line but applies to points on a curve.

• The slope is found via the derivative of the function at the chosen point.
• In implicit differentiation, you find \( \frac{dy}{dx} \) and evaluate it at the specified point.

In our exercise, the slope of the tangent line at point \((1,1)\) is determined by improving understanding of the implicit derivative.
Evaluating \( \frac{dy}{dx} \) by substituting the coordinates into the derived formula gives us \(-\frac{1}{4}\).
The negative sign indicates the tangent line decreases as it passes through \((1,1)\).
This understanding helps in visualizing how a curve behaves locally around any given point.