Problem 47 - Calculus and its applications

Question

Find \(\frac{d y}{d u}, \frac{d u}{d x},\) and \(\frac{d y}{d x}\) $$ y=u^{50} \text { and } u=4 x^{3}-2 x^{2} $$

Step-by-Step Solution

Answer

\( \frac{dy}{du} = 50u^{49} \), \( \frac{du}{dx} = 12x^2 - 4x \), \( \frac{dy}{dx} = 50(4x^3-2x^2)^{49}(12x^2-4x) \).

Step 1: Differentiate y with respect to u

Given the function \( y = u^{50} \). To find \( \frac{dy}{du} \), use the power rule which states that the derivative of \( u^n \) is \( n \cdot u^{n-1} \). Therefore, \( \frac{dy}{du} = 50u^{49} \).

Step 2: Differentiate u with respect to x

Given the function \( u = 4x^3 - 2x^2 \). Use the power rule to find \( \frac{du}{dx} \). The derivative of \( 4x^3 \) is \( 12x^2 \) and the derivative of \( -2x^2 \) is \( -4x \). Thus, \( \frac{du}{dx} = 12x^2 - 4x \).

Step 3: Find dy/dx using the chain rule

To find \( \frac{dy}{dx} \), use the chain rule \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \). Substitute the expressions from Step 1 and 2:\[\frac{dy}{dx} = (50u^{49})(12x^2 - 4x)\]Finally, substitute \( u = 4x^3 - 2x^2 \) into \( u^{49} \) to get the expression in terms of \( x \):\[\frac{dy}{dx} = 50(4x^3 - 2x^2)^{49}(12x^2 - 4x)\]

Key Concepts

Power Rule in Derivatives

When learning about derivatives, one of the most fundamental rules you'll come across is the power rule. This is a handy shortcut for differentiating functions where you have a variable raised to a power, like those in the form of \( u^n \). The power rule states that to differentiate \( u^n \) with respect to \( u \), you bring down the exponent as a coefficient and then subtract one from the exponent. This gives you the new expression \( n \cdot u^{n-1} \).

For example, with the function \( y = u^{50} \), applying the power rule means that the derivative \( \frac{dy}{du} \) becomes \( 50u^{49} \). This simplifies the process significantly, enabling you to quickly find derivatives of polynomial functions.

Using this rule is crucial in understanding how changes in variables affect the outcome of a function, and it's widely applied in calculus problems involving polynomials.

Understanding the Chain Rule

The chain rule is a powerful technique used when you need to differentiate composite functions. These are functions where one function is nested inside another. It may sound complex, but it simplifies nicely once you get used to it. The essence of the chain rule is that it allows you to take derivatives of these layers one at a time.

Consider a function \( y = f(g(x)) \), where \( g(x) \) is a function nested inside another function \( f \). The chain rule tells us that the derivative of \( y \) with respect to \( x \) is \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \). In this formula,

• \( \frac{dy}{du} \) is the derivative of the outer function \( f \) with respect to the inner function \( u \).
• \( \frac{du}{dx} \) is the derivative of the inner function \( g \) with respect to \( x \).

This process is akin to using linked chains; the change in one chain affects the next. For example, if you have \( y = u^{50} \) and \( u = 4x^3 - 2x^2 \), to find \( \frac{dy}{dx} \), simply find each derivative separately and multiply them. That's the beauty of the chain rule.

Implicit Differentiation Simplified

Sometimes, you'll encounter equations that involve multiple variables on both sides and are not in the standard form of \( y = f(x) \). In these cases, implicit differentiation becomes an invaluable tool. Instead of solving for \( y \) and then differentiating explicitly, you work with both variables side by side.

Implicit differentiation allows you to differentiate each part of the equation with respect to \( x \) directly. It often requires the application of the chain rule as you differentiate each term. For example, if \( y \) is implicitly defined in terms of \( x \) through an equation like \( F(x, y) = 0 \), you treat \( y \) as a function of \( x \).

To carry this out, differentiate every term with respect to \( x \), applying the product rule or chain rule as needed. Don't forget to multiply by \( \frac{dy}{dx} \) when differentiating terms involving \( y \). Solving these derivations will allow you to find relations and other derivatives that are critical in solving complex equations where it's not feasible or possible to express one variable explicitly as a function of another.