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Find the derivatives of the given functions. Assume that \(a, b, c,\) and \(k\) are constants. $$f(t)=3 t^{2}-4 t+1$$

Short Answer

Expert verified
The derivative of \( f(t) = 3t^2 - 4t + 1 \) is \( f'(t) = 6t - 4 \).

Step by step solution

01

Identify the function components

The function given is \( f(t) = 3t^2 - 4t + 1 \). It consists of three terms: \( 3t^2 \), \( -4t \), and \( 1 \). Each term will be differentiated individually.
02

Differentiate the first term

Differentiate \( 3t^2 \) using the power rule which states that \( \frac{d}{dt}(t^n) = nt^{n-1} \). Applying the rule, \( \frac{d}{dt}(3t^2) = 2 \cdot 3t^{2-1} = 6t \).
03

Differentiate the second term

Differentiate \( -4t \) using the power rule. Rewrite \( -4t \) as \( -4t^1 \) and differentiate: \( \frac{d}{dt}(-4t) = -4 \cdot 1t^{1-1} = -4 \).
04

Differentiate the third term

The last term is a constant, \( 1 \). The derivative of a constant is \( 0 \). Therefore, \( \frac{d}{dt}(1) = 0 \).
05

Combine the derivatives

Add together the derivatives of each term to get the derivative of the entire function: \( 6t - 4 + 0 = 6t - 4 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Rule
The Power Rule is a fundamental technique in calculus for finding derivatives of polynomial expressions. When a function is expressed in the form of a power, such as \( t^n \), the Power Rule makes differentiation straightforward. The rule states that the derivative of \( t^n \) is \( nt^{n-1} \).

Let's break it down with an example: for the function \( 3t^2 \), applying the Power Rule gives us \( \frac{d}{dt}(3t^2) = 2 \cdot 3t^{2-1} = 6t \). You multiply the exponent by the coefficient (3), reduce the exponent by one (2 becomes 1), and you get \( 6t \). This rule works for any real number power and is frequently used because many functions can be simplified or represented in this form.
  • The Power Rule is particularly useful when dealing with polynomials.
  • It simplifies the differentiation process when powers are involved.
  • Remembering that any constant multiplied by a power can be easily differentiated using this rule is key.
Differentiation
Differentiation is a core concept in calculus that involves finding the derivative of a function. The derivative represents the rate of change of the function, giving us information about how the function behaves as its input changes.

The process of differentiation applies various rules and techniques to transform a function into its derivative. With a function like \( f(t) = 3t^2 - 4t + 1 \), differentiation requires handling each term separately to find the derivative. First, identify each term of the function, and then apply the necessary rules, such as the Power Rule, to find the derivative.
  • In the given function, \( 3t^2 \) becomes \( 6t \), \(-4t\) becomes \(-4\), and the derivative of the constant \( 1 \) is \( 0 \).
  • Combining these, the derivative \( f'(t) \) equals \( 6t - 4 \).
Differentiation is all about understanding how different function components contribute to the overall rate of change.
Calculus Concepts
Calculus is the mathematical study of change, and its core concepts are differentiation and integration. Differentiation, as we discussed, provides insight into how functions change, and it's essential in a variety of fields, from physics to economics.

Within calculus, various rules and methods are employed to solve problems related to changing quantities. In the case of the function \( f(t) = 3t^2 - 4t + 1 \), applying calculus concepts allows us to determine how the function's value changes with respect to \( t \). This is vital, as understanding these changes can help predict behavior and outcomes in real-world applications.
  • Calculus concepts stress the importance of both learning the rules and understanding their applications.
  • These concepts build a foundation for analyzing and solving complex problems involving rates of change.
  • Mastery of calculus opens doors to exploring dynamic systems and phenomena in various domains.
To effectively harness the tools of calculus, one must connect conceptual understanding with practical computation, thereby gaining a robust analytical skill set.

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Most popular questions from this chapter

At pressure \(P\) atmospheres, a certain fraction \(f\) of a gas decomposes. The quantities \(P\) and \(f\) are related, for some positive constant \(K,\) by the equation $$ \frac{4 f^{2} P}{1-f^{2}}=K $$ (a) Find \(d f / d P\) (b) Show that \(d f / d P<0\) always. What does this mean in practical terms?

The number of years, \(T\), it takes an investment of \(\$ 1000\) to grow to \(\$ F\) in an account which pays \(5 \%\) interest compounded continuously is given by $$T=g(F)=20 \ln (0.001 F)$$ Find \(g(5000)\) and \(g^{\prime}(5000) .\) Give units with your answers and interpret them in terms of money in the account.

(a) Find tanh 0 (b) For what values of \(x\) is tanh \(x\) positive? Negative? Explain your answer algebraically. (c) On what intervals is tanh \(x\) increasing? Decreasing? Use derivatives to explain your answer. (d) Find \(\lim _{x \rightarrow \infty} \tanh x\) and \(\lim _{x \rightarrow-\infty} \tanh x .\) Show this information on a graph. (e) Does tanh \(x\) have an inverse? Justify your answer using derivatives.

A wave travels along a string that is joined to a thicker rope. The wave both reflects back along the string and is transmitted to the rope. For positive constants \(k_{1}, k_{2}, w\) and time \(t,\) the wave along the string, given by \(x<0,\) is $$f(x)=\cos \left(k_{1} x-w t\right)+R \cos \left(-k_{1} x-w t\right)$$and the wave along the rope, given by \(x>0,\) is $$ g(x)=T \cos \left(k_{2} x-w t\right)$$ For every value of \(t,\) the two waves have the same tangent line at \(x=0,\) so they have the same value and same slope at \(x=0 .\) Use this fact to show that $$R=\frac{k_{1}-k_{2}}{k_{1}+k_{2}} \quad \text { and } \quad T=\frac{2 k_{1}}{k_{1}+k_{2}}$$ These amplitudes are called the reflection coefficient, \(R\) and the transmission coefficient, \(T\).

Explain what is wrong with the statement. If \(w(x)=\ln \left(1+x^{4}\right)\) then \(w^{\prime}(x)=1 /\left(1+x^{4}\right).\)

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