Chain Rule How To Find Dw/dt Examples And Explanation

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The chain rule is a fundamental concept in calculus that allows us to differentiate composite functions. In single-variable calculus, the chain rule states that the derivative of a composite function is the product of the derivative of the outer function evaluated at the inner function and the derivative of the inner function. This concept extends to multivariable calculus, where we can use the chain rule to find the rate of change of a function with respect to a variable when the function depends on intermediate variables, which in turn depend on the variable of interest. In this article, we will explore how to apply the chain rule to find dw/dt for various functions w, where w depends on x, y, and z, and x, y, and z are functions of t. We'll delve into specific examples, providing a step-by-step guide to understanding and applying this crucial calculus concept.

The multivariable chain rule is an extension of the single-variable chain rule, designed to handle functions that depend on multiple variables. In the context of our problem, we have a function w(x, y, z), where x, y, and z are themselves functions of t. The chain rule allows us to find the derivative of w with respect to t, denoted as dw/dt. This derivative represents the rate of change of w as t changes.

To apply the chain rule, we consider the partial derivatives of w with respect to each intermediate variable (x, y, and z) and the derivatives of these intermediate variables with respect to t. The formula for dw/dt is given by:

dw/dt = (∂w/∂x)(dx/dt) + (∂w/∂y)(dy/dt) + (∂w/∂z)(dz/dt)

This formula tells us that the total change in w with respect to t is the sum of the changes in w due to changes in x, y, and z, each weighted by the rate at which x, y, and z change with respect to t. This formula is the cornerstone of solving the problems presented in this article. Understanding this formula is crucial for anyone studying multivariable calculus and its applications.

In this problem, we are given the function w = 5x²y³z⁴, where x = t², y = t³, and z = t⁵. Our goal is to find dw/dt using the chain rule. This involves finding the partial derivatives of w with respect to x, y, and z, as well as the derivatives of x, y, and z with respect to t. We will then substitute these values into the chain rule formula to obtain dw/dt. This is a classic application of the chain rule in multivariable calculus, and it demonstrates how to handle polynomial functions within the context of this rule.

First, we find the partial derivatives of w:

  • ∂w/∂x = 10xy³z⁴
  • ∂w/∂y = 15x²y²z⁴
  • ∂w/∂z = 20x²y³z³

Next, we find the derivatives of x, y, and z with respect to t:

  • dx/dt = 2t
  • dy/dt = 3t²
  • dz/dt = 5t⁴

Now, we substitute these into the chain rule formula:

dw/dt = (10xy³z⁴)(2t) + (15x²y²z⁴)(3t²) + (20x²y³z³)(5t⁴)

Finally, we substitute x = t², y = t³, and z = t⁵ into the equation:

dw/dt = (10(t²)(t³)³(t⁵)⁴)(2t) + (15(t²)²(t³)²(t⁵)⁴)(3t²) + (20(t²)²(t³)²(t⁵)³)(5t⁴)

Simplifying the expression, we get:

dw/dt = (10t²t⁹t²⁰)(2t) + (15t⁴t⁶t²⁰)(3t²) + (20t⁴t⁹t¹⁵)(5t⁴)

dw/dt = 20t³² + 45t³² + 100t³²

dw/dt = 165t³²

Therefore, the derivative of w with respect to t is 165t³². This result shows how the chain rule allows us to combine the rates of change from different variables to find the overall rate of change of the function w.

In this problem, we are given the function w = ln(3x² - 2y + 4z³), where x = t^(1/2), y = t^(2/3), and z = t^(-2). We need to find dw/dt using the chain rule. This problem involves a logarithmic function, which adds a layer of complexity to the differentiation process. We will again find the partial derivatives of w with respect to x, y, and z, and the derivatives of x, y, and z with respect to t. Then, we will substitute these values into the chain rule formula to obtain dw/dt. This example is particularly useful for understanding how to apply the chain rule when dealing with logarithmic and power functions.

First, we find the partial derivatives of w:

  • ∂w/∂x = (6x) / (3x² - 2y + 4z³)
  • ∂w/∂y = (-2) / (3x² - 2y + 4z³)
  • ∂w/∂z = (12z²) / (3x² - 2y + 4z³)

Next, we find the derivatives of x, y, and z with respect to t:

  • dx/dt = (1/2)t^(-1/2)
  • dy/dt = (2/3)t^(-1/3)
  • dz/dt = -2t^(-3)

Now, we substitute these into the chain rule formula:

dw/dt = [(6x) / (3x² - 2y + 4z³)][(1/2)t^(-1/2)] + [(-2) / (3x² - 2y + 4z³)][(2/3)t^(-1/3)] + [(12z²) / (3x² - 2y + 4z³)][-2t^(-3)]

Finally, we substitute x = t^(1/2), y = t^(2/3), and z = t^(-2) into the equation:

dw/dt = [(6t^(1/2)) / (3(t^(1/2))² - 2t^(2/3) + 4(t(-2))³)][(1/2)t(-1/2)] + [(-2) / (3(t^(1/2))² - 2t^(2/3) + 4(t(-2))³)][(2/3)t(-1/3)] + [(12(t^(-2))²) / (3(t^(1/2))² - 2t^(2/3) + 4(t(-2))³)][-2t(-3)]

Simplifying the expression, we get:

dw/dt = [3t(1/2)t(-1/2) / (3t - 2t^(2/3) + 4t^(-6))] + [-4/3 t^(-1/3) / (3t - 2t^(2/3) + 4t^(-6))] + [-24t(-4)t(-3) / (3t - 2t^(2/3) + 4t^(-6))]

dw/dt = [3 / (3t - 2t^(2/3) + 4t^(-6))] + [-4/3 t^(-1/3) / (3t - 2t^(2/3) + 4t^(-6))] + [-24t^(-7) / (3t - 2t^(2/3) + 4t^(-6))]

Combining the terms, we have:

dw/dt = [3 - (4/3)t^(-1/3) - 24t^(-7)] / [3t - 2t^(2/3) + 4t^(-6)]

Therefore, the derivative of w with respect to t is [3 - (4/3)t^(-1/3) - 24t^(-7)] / [3t - 2t^(2/3) + 4t^(-6)]. This result demonstrates how the chain rule can be applied to more complex functions, such as those involving logarithms and fractional exponents.

In this problem, we are tasked with finding dw/dt for the function w = 5 cos(xy) - sin(xz), where x = 1/t, y = t, and z = t³. This problem introduces trigonometric functions into the mix, which require careful application of differentiation rules. As before, we will find the partial derivatives of w with respect to x, y, and z, and the derivatives of x, y, and z with respect to t. Then, we will substitute these values into the chain rule formula to obtain dw/dt. This example is particularly helpful for understanding how to apply the chain rule when trigonometric functions are involved.

First, we find the partial derivatives of w:

  • ∂w/∂x = -5y sin(xy) - z cos(xz)
  • ∂w/∂y = -5x sin(xy)
  • ∂w/∂z = -x cos(xz)

Next, we find the derivatives of x, y, and z with respect to t:

  • dx/dt = -1/t²
  • dy/dt = 1
  • dz/dt = 3t²

Now, we substitute these into the chain rule formula:

dw/dt = -5y sin(xy) - z cos(xz) + -5x sin(xy) + -x cos(xz)

Finally, we substitute x = 1/t, y = t, and z = t³ into the equation:

dw/dt = -5t sin((1/t)t) - t³ cos((1/t)t³) + -5(1/t) sin((1/t)t) + -(1/t) cos((1/t)t³)

Simplifying the expression, we get:

dw/dt = -5t sin(1) - t³ cos(t²) + [-5(1/t) sin(1)] + -(1/t) cos(t²)

dw/dt = [5t sin(1) + t³ cos(t²)]/t² - (5/t) sin(1) - 3t cos(t²)

dw/dt = (5 sin(1))/t + t cos(t²) - (5 sin(1))/t - 3t cos(t²)

dw/dt = -2t cos(t²)

Therefore, the derivative of w with respect to t is -2t cos(t²). This result showcases how the chain rule can effectively handle trigonometric functions and their derivatives within a multivariable context.

In this final problem, we are given the function w = √(1 + x - 2yz⁴x), where x = ln t, y = t, and z = 4t. Our objective is to find dw/dt using the chain rule. This problem involves a square root function, which requires careful application of the power rule and the chain rule. As in the previous examples, we will find the partial derivatives of w with respect to x, y, and z, and the derivatives of x, y, and z with respect to t. Then, we will substitute these values into the chain rule formula to obtain dw/dt. This example is valuable for understanding how to apply the chain rule when dealing with radical functions and logarithmic functions.

First, we find the partial derivatives of w:

w = (1 + x - 2yz⁴x)^(1/2)

  • ∂w/∂x = (1/2)(1 + x - 2yz⁴x)^(-1/2)(1 - 2yz⁴)
  • ∂w/∂y = (1/2)(1 + x - 2yz⁴x)^(-1/2)(-2z⁴x)
  • ∂w/∂z = (1/2)(1 + x - 2yz⁴x)^(-1/2)(-8yz³x)

Next, we find the derivatives of x, y, and z with respect to t:

  • dx/dt = 1/t
  • dy/dt = 1
  • dz/dt = 4

Now, we substitute these into the chain rule formula:

dw/dt = (1/2)(1 + x - 2yz⁴x)^(-1/2)(1 - 2yz⁴) + (1/2)(1 + x - 2yz⁴x)^(-1/2)(-2z⁴x) + (1/2)(1 + x - 2yz⁴x)^(-1/2)(-8yz³x)

Finally, we substitute x = ln t, y = t, and z = 4t into the equation:

dw/dt = (1/2)(1 + ln t - 2t(4t)⁴ln t)^(-1/2)(1 - 2t(4t)⁴) + (1/2)(1 + ln t - 2t(4t)⁴ln t)^(-1/2)(-2(4t)⁴ln t) + (1/2)(1 + ln t - 2t(4t)⁴ln t)^(-1/2)(-8t(4t)³ln t)

Simplifying the expression, we get:

dw/dt = (1/2)(1 + ln t - 512t⁵ln t)^(-1/2)(1 - 512t⁵) + [(1/2)(1 + ln t - 512t⁵ln t)^(-1/2)(-512t⁴ln t)] + (1/2)(1 + ln t - 512t⁵ln t)^(-1/2)(-512t⁴ln t)

dw/dt = (1/2)(1 + ln t - 512t⁵ln t)^(-1/2) [(1 - 512t⁵)/t - 512t⁴ln t - 2048t⁴ln t]

dw/dt = (1/2)(1 + ln t - 512t⁵ln t)^(-1/2) [(1 - 512t⁵)/t - 2560t⁴ln t]

Therefore, the derivative of w with respect to t is (1/2)(1 + ln t - 512t⁵ln t)^(-1/2) [(1 - 512t⁵)/t - 2560t⁴ln t]. This final example demonstrates how the chain rule can be applied to a function involving a square root, logarithmic functions, and polynomial terms, highlighting the versatility and power of the chain rule in multivariable calculus.

In this article, we have explored the application of the chain rule in multivariable calculus to find dw/dt for various functions. We have demonstrated how to apply the chain rule to functions involving polynomials, logarithms, trigonometric functions, and radicals. The key to successfully applying the chain rule is to correctly identify the intermediate variables and their relationships, compute the partial derivatives and derivatives with respect to t, and substitute these into the chain rule formula. These examples provide a comprehensive guide to understanding and applying the chain rule in a variety of contexts, making it an indispensable tool for anyone studying calculus and its applications. The chain rule is not just a mathematical formula; it is a fundamental concept that helps us understand how rates of change propagate through interconnected systems, making it a valuable tool in various fields of science and engineering.