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Magazine Chapter 29: Problem 23
Find the differential coefficient of $$ y=\frac{2}{\left(2 t^{3}-5\right)^{4}} $$
Short Answer
Expert verified
The differential coefficient is \( \frac{dy}{dt} = -48t^2(2t^3-5)^{-5} \).
Step by step solution
01
Recognize the Chain Rule
To find the derivative of a function that is a composition of functions, we can apply the chain rule. Here, our function is a composite function, with the outer function being a power of -4 and the inner function being a polynomial.
02
Express the Function for Differentiation
Given the function \( y = \frac{2}{(2t^3 - 5)^4} \), we can rewrite it as \( y = 2(2t^3 - 5)^{-4} \) to make differentiation easier.
03
Differentiate the Outer Function
Differentiate the outer function \( u^{-4} \) with respect to \( u \), which results in \(-4u^{-5}\). Apply this derivative to the inner function \( u = 2t^3 - 5 \).
04
Differentiate the Inner Function
Now differentiate the inner function \( 2t^3 - 5 \) with respect to \( t \). The derivative of \( 2t^3 \) is \( 6t^2 \), and the derivative of \( -5 \) is \( 0 \). So the derivative of the inner function is \( 6t^2 \).
05
Apply the Chain Rule
Applying the chain rule, multiply the derivative of the outer function by the derivative of the inner function:\[ \frac{dy}{dt} = 2 \cdot (-4(2t^3-5)^{-5}) \cdot (6t^2) \].
06
Simplify the Expression
Simplify the expression by calculating \(-4 \times 2 \times 6t^2\) to get \(-48t^2\). So, the differential coefficient is:\[ \frac{dy}{dt} = -48t^2(2t^3-5)^{-5} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivatives
Derivatives are fundamental in calculus as they reveal how a function changes. This is often related to finding the slope of the tangent line at any point on a curve.
- A derivative provides the rate of change or the instantaneous rate of change, vital in various scientific fields.
- If you have a function, its derivative is usually represented as \( \frac{dy}{dt} \) or \( f'(x) \).
- To compute a derivative at a certain point gives us the value of how steep the function is at that point.
Chain Rule
The chain rule is a powerful technique used in calculus to differentiate composite functions. That is, functions within functions. When dealing with a composite function, the derivative of the "outer" function is taken first, then multiplied by the derivative of the "inner" function.
- For a function \( f(g(x)) \), the chain rule states that the derivative \( f'(x) \) is \( f'(g(x)) \cdot g'(x) \).
- This method is essential when functions are not straightforward, especially when differentiating trigonometric, logarithmic, or exponential functions within other functions.
- Remember to identify both components of the composition to apply the chain rule correctly: what is being done to the variable \( x \), and what is happening inside that operation.
Differential Equations
Differential equations involve derivatives and are used to describe various phenomena dynamically. They play a crucial part in modeling the physical world.
- A differential equation often relates a function to its derivatives, expressing how the function evolves with time or space.
- They can be ordinary (ODEs) involving functions of a single variable, or partial (PDEs) involving several variables.
- Solutions to these equations can predict behavior in systems ranging from simple harmonic motion to complex economic models.
Polynomial Functions
Polynomial functions are a cornerstone in mathematics, consisting of variables raised to whole number powers, combined using addition, subtraction, or multiplication.
- These functions are written in the standard form \( a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \), where each term \( a_i \) is a constant coefficient.
- They are straightforward to differentiate, as the power rule applies, reducing each term's power by one and multiplying by the original power.
- Polynomials of degree two or higher provide simple models of nonlinear relationships, capturing more complex behaviors than linear functions.
