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Chapter 10: Problem 2

Evaluate the derivatives of the functions at the given value: (a) \(y=2 t+9+\mathrm{e}^{t / 2} \quad t=1\) (b) \(y=\frac{t^{2}-4 t+6}{3} \quad t=2\) (c) \(y=\sin t+\cos t \quad t=1\) (d) \(y=3 \mathrm{e}^{2 t}-2 \sin \left(\frac{t}{2}\right) \quad t=0\) (e) \(y=5 \tan (2 x)+\frac{1}{\mathrm{e}^{2 x}} \quad x=0.5\) (f) \(y=3 \ln t+\sin (3 t) \quad t=0.25\)

Short Answer

Expert verified
Evaluate derivatives at given values for precise solutions.

Step by step solution

01

Differentiate (a)

The function is given by \( y = 2t + 9 + e^{t/2} \). To find the derivative, use the power rule and chain rule.- The derivative of \( 2t \) is 2.- The derivative of the constant 9 is 0.- The derivative of \( e^{t/2} \) is \( \frac{1}{2} e^{t/2} \), using the chain rule.Thus, the derivative is given by \[ y' = 2 + \frac{1}{2} e^{t/2}. \]
02

Evaluate at t=1 for (a)

Substitute \( t = 1 \) into the derivative:\[ y'(1) = 2 + \frac{1}{2} e^{1/2}. \]This yields the derivative value at \( t = 1 \).
03

Differentiate (b)

The function is \( y = \frac{t^2 - 4t + 6}{3} \). Simplify the derivative by first distributing 1/3:\[ y = \frac{1}{3}t^2 - \frac{4}{3}t + 2. \]Differentiate each term:- The derivative of \( \frac{1}{3}t^2 \) is \( \frac{2}{3}t \).- The derivative of \( -\frac{4}{3}t \) is \( -\frac{4}{3} \).Thus, the derivative is:\[ y' = \frac{2}{3}t - \frac{4}{3}. \]
04

Evaluate at t=2 for (b)

Substitute \( t = 2 \) into the derivative:\[ y'(2) = \frac{2}{3}(2) - \frac{4}{3}. \]Calculate the result to obtain the derivative at \( t = 2 \).
05

Differentiate (c)

The function is \( y = \sin t + \cos t \). Find its derivative using basic derivatives of trigonometric functions:- The derivative of \( \sin t \) is \( \cos t \).- The derivative of \( \cos t \) is \( -\sin t \).Thus, the derivative is:\[ y' = \cos t - \sin t. \]
06

Evaluate at t=1 for (c)

Substitute \( t = 1 \) into the derivative:\[ y'(1) = \cos(1) - \sin(1). \]This evaluates the derivative at \( t = 1 \).
07

Differentiate (d)

The function is \( y = 3e^{2t} - 2 \sin\left(\frac{t}{2}\right) \). Apply the chain rule for each term:- The derivative of \( 3e^{2t} \) is \( 6e^{2t} \) because \( e^{2t} \) differentiates to \( 2e^{2t} \) and multiplied by 3.- The derivative of \( 2 \sin\left(\frac{t}{2}\right) \) is \( -\cos\left(\frac{t}{2}\right) \) due to the chain rule.Thus, the derivative is:\[ y' = 6e^{2t} - \cos\left(\frac{t}{2}\right). \]
08

Evaluate at t=0 for (d)

Substitute \( t = 0 \) into the derivative:\[ y'(0) = 6e^{0} - \cos(0). \]Calculate the result to find the derivative at \( t = 0 \).
09

Differentiate (e)

The function is \( y = 5 \tan(2x) + \frac{1}{e^{2x}} \). Differentiate each term:- The derivative of \( 5 \tan(2x) \) is \( 10 \sec^2(2x) \) using the chain rule.- The derivative of \( \frac{1}{e^{2x}} \) is \( -2e^{-2x} \) using the chain rule.Thus, the derivative is:\[ y' = 10 \sec^2(2x) - 2e^{-2x}. \]
10

Evaluate at x=0.5 for (e)

Substitute \( x = 0.5 \) into the derivative:\[ y'(0.5) = 10 \sec^2(1) - 2 \cdot e^{-1}. \]This gives the derivative value at \( x = 0.5 \).
11

Differentiate (f)

The function is \( y = 3 \ln t + \sin(3t) \). Differentiating both terms gives:- The derivative of \( 3 \ln t \) is \( \frac{3}{t} \).- The derivative of \( \sin(3t) \) is \( 3 \cos(3t) \) using the chain rule.Thus, the derivative is:\[ y' = \frac{3}{t} + 3 \cos(3t). \]
12

Evaluate at t=0.25 for (f)

Substitute \( t = 0.25 \) into the derivative:\[ y'(0.25) = \frac{3}{0.25} + 3 \cos(0.75). \]This calculates the derivative at \( t = 0.25 \).

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Key Concepts

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

Chain Rule
The Chain Rule is an essential tool in calculus when it comes to taking derivatives of composite functions. It allows us to differentiate a function inside another function. Think of it as peeling an onion layer by layer, until you get to the center.

Here's a simple way to understand how it works:
  • If you have a function like \( e^{t/2} \), you're really dealing with two functions: the exponential function \( e^u \), where \( u = t/2 \).
  • To find the derivative, you first differentiate \( e^u \) regarding \( u \), which is just \( e^u \), and then differentiate \( u \) as \( t/2 \), which gives \( 1/2 \).
  • Finally, you multiply these two derivatives together to get \( \, (1/2)e^{t/2} \).
This rule is particularly useful when dealing with functions like \( e^{2t} \), \( an(2x) \), and \( \sin(3t) \), as seen in the provided examples.
Evaluation of Derivatives
Often, after calculating the derivative of a function, you'll need to evaluate it at a certain point. This is simply a matter of substitution.

Here's how you do it:
  • Suppose you've already determined the derivative, like in example (b) where \( y' = \frac{2}{3}t - \frac{4}{3} \).
  • If you want to find out the rate of change at a specific point \( t = 2 \), you just plug in the value into the derivative: \( y'(2) = \frac{2}{3} \times 2 - \frac{4}{3} \).
  • This evaluates to a numeric value showing the slope or rate of change at that exact \( t \) value.
The same principle applies to all types of functions鈥攚hether simple or composed鈥攎aking it a universal step in solving derivative problems.
Power Rule
The Power Rule is possibly one of the easiest and most frequently used rules for differentiation. It says that if you have a function \( x^n \), the derivative \( \frac{d}{dx}(x^n) \) is \( nx^{n-1} \).

In simpler terms:
  • You bring down the exponent as a coefficient.
  • Then, you reduce the original exponent by one.
  • For example, differentiating \( \frac{1}{3}t^2 \) yields \( \frac{2}{3}t \).
Whenever you see a polynomial, you can immediately apply the Power Rule to each component. This makes solving derivatives faster and simpler, like a math shortcut based on straightforward rules.
Trigonometric Derivatives
Trigonometric functions like sine and cosine have their own special derivatives. These derivatives are at the very core of calculus when dealing with periodic functions.

Let鈥檚 lighten up the process:
  • The derivative of \( \sin t \) is \( \cos t \).
  • The derivative of \( \cos t \) is \( -\sin t \).
  • The derivative of \( \tan t \) is \( \sec^2 t \), and with the chain rule, \( \tan(2x) \) becomes \( 10 \sec^2(2x) \).
Given how often these appear in exercises, knowing them by heart can speed up your problem-solving. These rules apply directly to find changes in angles and cycles, proving immensely useful in fields inclusive of engineering and physics.

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Most popular questions from this chapter

The function \(g(t)\) is defined by $$ g(t)= \begin{cases}0 & t<0 \\ t^{2} & 0 \leqslant t \leqslant 3 \\ 2 t+3 & 34\end{cases} $$ (a) Sketch \(g\). (b) State any points of discontinuity. (c) Find, if they exist, (i) \(\lim _{t \rightarrow 3} g\) (ii) \(\lim _{t \rightarrow 4} g\) (iii) \(\lim _{t \rightarrow 4^{-}} g\)

Find the rate of change of \(y(x)=\frac{x}{x+3}\) at \(x=3\) considering the interval \([3,3+\delta x]\).

The function, \(f(t)\), is defined by $$ f(t)= \begin{cases}1 & 0 \leqslant t \leqslant 2 \\ 2 & 23\end{cases} $$ Sketch a graph of \(f(t)\) and state the following limits if they exist: (a) \(\lim _{t \rightarrow 1.5} f\) (b) \(\lim _{t \rightarrow 2^{+}} f\) (c) \(\lim _{t \rightarrow 3} f\) (d) \(\lim _{t \rightarrow 0^{+}} f\) (e) \(\lim _{t \rightarrow 3^{-}} f\)

A function, \(y\), is such that \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) is constant. What can you say about the function, \(y\) ?

Differentiate the following functions: (a) \(y=4 x^{3}-5 x^{2}\) (b) \(y=3 \sin (5 t)+2 \mathrm{e}^{4 t}\) (c) \(y=\sin (4 t)+3 \cos (2 t)-t\) (d) \(y=\tan (3 z)\) (e) \(y=2 \mathrm{e}^{3 t}+17-4 \sin (2 t)\) (f) \(y=\frac{1}{t^{3}}+\frac{\cos 5 t}{2}\) (g) \(y=\frac{2 w^{3}}{3}+\frac{\mathrm{e}^{4 w}}{2}\) (h) \(y=\sqrt{x}+\ln (\sqrt{x})\). [Hint: \(\sqrt{x}=x^{1 / 2}\), and use the laws of logarithms.]
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