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Chapter 5: Problem 2

Compute the derivatives of a. \(\quad P(t)=e^{\left(t^{2}\right)}\) b. \(P(t)=\ln \left(t^{2}\right)\) c. \(P(t)=\left(e^{t}\right)^{2}\) d. \(P(t)=e^{(2 \ln t)}\) e. \(P(t)=\ln \left(e^{3 t}\right)\) f. \(\quad P(t)=\sqrt{e^{t}}\) g. \(\quad P(t)=e^{5}\) h. \(P(t)=\ln (\sqrt{t})\) i. \(\quad P(t)=e^{t+1}\)

Short Answer

Expert verified
a. 2te^{t^2}, b. \frac{2}{t}, c. 2e^{2t}, d. 2t, e. 3, f. \frac{1}{2}e^{t/2}, g. 0, h. \frac{1}{2t}, i. e^{t+1}.

Step by step solution

01

Differentiate Exponential Function

To differentiate \( P(t) = e^{t^2} \), use the chain rule. Let \( u = t^2 \). Then \( \frac{du}{dt} = 2t \). The derivative is given by \( \frac{dP}{dt} = e^{u} \cdot \frac{du}{dt} = e^{t^2} \cdot 2t = 2te^{t^2} \).
02

Differentiate Logarithmic Function

For \( P(t) = \ln(t^2) \), use the chain rule. Let \( u = t^2 \) so \( \frac{du}{dt} = 2t \). The derivative is \( \frac{dP}{dt} = \frac{1}{u} \cdot \frac{du}{dt} = \frac{1}{t^2} \cdot 2t = \frac{2}{t} \).
03

Differentiate Power of Exponential

For \( P(t) = (e^t)^2 \), rewrite as \( P(t) = e^{2t} \). Differentiate using the chain rule, \( \frac{dP}{dt} = e^{2t} \cdot 2 = 2e^{2t} \).
04

Simplify and Differentiate Exponential-Logarithm

Given \( P(t) = e^{2 \ln t} \), simplify as \( P(t) = t^2 \). The derivative \( \frac{dP}{dt} = 2t \).
05

Differentiate Logarithm of Exponential

For \( P(t) = \ln(e^{3t}) \), use \( P(t) = 3t \). Thus, \( \frac{dP}{dt} = 3 \).
06

Differentiate Square Root of Exponential

Given \( P(t) = \sqrt{e^t} \), rewrite as \( P(t) = e^{t/2} \). Differentiate using the chain rule: \( \frac{dP}{dt} = e^{t/2} \cdot \frac{1}{2} = \frac{1}{2}e^{t/2} \).
07

Differentiate Constant Exponential

For \( P(t) = e^5 \), which is a constant, its derivative is \( 0 \).
08

Differentiate Logarithm of Square Root

For \( P(t) = \ln(\sqrt{t}) \), rewrite as \( P(t) = \frac{1}{2} \ln(t) \). Thus, \( \frac{dP}{dt} = \frac{1}{2} \cdot \frac{1}{t} = \frac{1}{2t} \).
09

Differentiate Sum of Exponential with Constant

For \( P(t) = e^{t+1} \), rewrite as \( P(t) = e^{t} \cdot e \). The derivative is \( \frac{dP}{dt} = e^{t} \cdot e = e^{t+1} \).

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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 a powerful tool in calculus that helps us differentiate composite functions. A composite function is like a function within another function, and the chain rule allows us to break down this complexity.
For example, if you have a function of the form \( P(t) = e^{t^2} \), you can think of it as \( P(t) = f(g(t)) \) where \( f(u) = e^u \) and \( g(t) = t^2 \). The chain rule states that the derivative of a composite function \( f(g(t)) \) is \( f'(g(t)) \cdot g'(t) \).
  • First, compute the derivative of the outer function \( f(u) = e^u \) as \( f'(u) = e^u \).
  • Next, compute the derivative of the inner function \( g(t) = t^2 \) as \( g'(t) = 2t \).
  • Finally, combine these using the chain rule: \( rac{dP}{dt} = e^{t^2} \cdot 2t = 2te^{t^2} \).
Logarithmic Differentiation
Logarithmic differentiation is especially useful when dealing with functions involving products, quotients, or powers. Applying this technique often makes the differentiation process easier.
Consider the function \( P(t) = \ln(t^2) \). To find its derivative, apply the rule for differentiating a logarithmic function: \( \frac{d}{dt}\ln(u) = \frac{1}{u} \cdot \frac{du}{dt} \).
  • Set \( u = t^2 \) leading to \( \frac{du}{dt} = 2t \).
  • Substitute into the differentiation formula: \( \frac{dP}{dt} = \frac{1}{t^2} \cdot 2t = \frac{2}{t} \).
Logarithmic differentiation can greatly simplify the calculation if you encounter complex expressions.
Exponential Functions
Exponential functions like \( e^t \) show up a lot in math and science because of their unique properties. They grow very fast and have the convenient property that their derivative is the same exponential function multiplied by the derivative of the exponent.
When differentiating, remember:
  • If \( P(t) = e^{u(t)} \), then \( \frac{dP}{dt} = e^{u(t)} \cdot \frac{du}{dt} \).
Consider \( P(t) = (e^t)^2 = e^{2t} \). Here, the function already simplifies to a standard form:
  • The derivative is straightforward: \( \frac{dP}{dt} = e^{2t} \cdot 2 = 2e^{2t} \).
Exponential functions can also involve other operations, such as powers and products, which you'll often handle using rules like the chain rule.
Constant Functions
A function is constant if it does not change as its variable changes. A classic example is \( P(t) = e^5 \). Regardless of what \( t \) is, the value of the function remains \( e^5 \).
The derivative of a constant function is always zero because the function is constant鈥攖here鈥檚 no change with respect to \( t \).
  • Thus, \( \frac{d}{dt}[e^5] = 0 \).
Understanding constant functions is important as they often represent fixed values in a problem or simplify expressions during differentiation.

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Most popular questions from this chapter

Draw the graphs of $$y(t)=e^{t} \quad \text { and } \quad p(t)=1+t+\frac{t^{2}}{2}+\frac{t^{3}}{6}+\frac{t^{4}}{24}$$ Set the domain and range to \(-1 \leq t \leq 2\). and \(0 \leq y \leq 8\).

Identify the errors in the following derivative computations. \(\begin{array}{l}\text { a. }\left[\left(t^{4}+t^{-1}\right)^{7}\right]^{\prime} \\ & 7\left(t^{4}+t^{-1}\right)^{6}\left[t^{4}+t^{-1}\right]^{\prime} \\ & 7\left(t^{4}+t^{-1}\right)^{6}\left[t^{4}\right]^{\prime}+\left[t^{-1}\right]^{\prime} \\\ & 7\left(t^{4}+t^{-1}\right)^{6} 4 t^{3}+(-1) t^{-2} \\ & 28 t^{3}\left(t^{4}+t^{-1}\right)^{6}-t^{-2}\end{array}\) \(\begin{array}{l}\text { b. }\left[5 t^{7}+7 t^{-5}\right]^{\prime} \\\ {\left[5 t^{7}\right]^{\prime}+\left[7 t^{-5}\right]^{\prime}} \\ & 5\left[t^{7}\right]^{\prime}+7\left[t^{-5}\right]^{\prime} \\ & 5 \times 7 t^{6}+7 \times(-5) t^{-4} \\ & 35\left(t^{6}-t^{-4}\right)\end{array}\) c. \(\left[10 t^{8}+8 e^{5 t}\right]^{\prime}\) \(\left[10 t^{8}\right]^{\prime}+\left[8 e^{5 t}\right]^{\prime}\) \(10\left[t^{8}\right]^{\prime}+8\left[e^{5 t}\right]^{\prime}\) \(10 \times 8 t^{7}+8 \times 5 e^{4 t}\) \(40\left(2 t^{7}+e^{4 t}\right)\)

Estimate the slope of the tangent to the graph of $$y=\log _{10} x$$ at the point \(\left(3, \log _{10} 3\right)\) correct to three decimal digits.

Let \(y(x)=e^{x}\). Compute \(y^{\prime}(x), y^{\prime \prime}(x)=\left(y^{\prime}\right)^{\prime},\) and \(y^{\prime \prime \prime}(x)\).

Draw the graphs of $$y_{1}(t)=e^{t} \quad y_{2}(t)=t^{2} \quad y_{3}(t)=t^{3} \quad y_{4}(t)=t^{4} \quad-1 \leq t \leq 5$$ The graphs are close together near \(t=0\) and increase as \(t\) increases. Which one grows the most as \(t\) increases? Expand the domain and range to \(-1 \leq t \leq 10,0 \leq y \leq 25,000,\) and answer the same question.
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