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Chapter 5: Problem 2

Differentiate each of the following: a. \(y=e^{3 x}\) b. \(s=e^{3 t-5}\) c. \(y=2 e^{10 t}\) d. \(y=e^{-3 x}\) e. \(y=e^{5-6 x+x^{2}}\) f. \(y=e^{\sqrt{x}}\)

Short Answer

Expert verified
a. \(3e^{3x}\), b. \(3e^{3t-5}\), c. \(20e^{10t}\), d. \(-3e^{-3x}\), e. \(e^{5-6x+x^2}(2x-6)\), f. \(\frac{e^{\sqrt{x}}}{2\sqrt{x}}\).

Step by step solution

01

Differentiate Part a

Given function is \(y = e^{3x}\). The derivative of \(e^{u}\) with respect to \(x\) is \(e^{u} \cdot \frac{du}{dx}\), where \(u = 3x\). Therefore, \(\frac{du}{dx} = 3\). So, the derivative \(\frac{dy}{dx} = e^{3x} \cdot 3 = 3e^{3x}\).
02

Differentiate Part b

Given function is \(s = e^{3t-5}\). Here, the exponent is \(u = 3t - 5\). Therefore, \(\frac{du}{dt} = 3\). The derivative of \(s\) with respect to \(t\) is \(\frac{ds}{dt} = e^{3t-5} \cdot 3 = 3e^{3t-5}\).
03

Differentiate Part c

Given function is \(y = 2e^{10t}\). Here, \(u = 10t\), thus \(\frac{du}{dt} = 10\). The derivative \(\frac{dy}{dt} = 2e^{10t} \cdot 10 = 20e^{10t}\).
04

Differentiate Part d

Given function is \(y = e^{-3x}\). Here, \(u = -3x\), so \(\frac{du}{dx} = -3\). The derivative \(\frac{dy}{dx} = e^{-3x} \cdot (-3) = -3e^{-3x}\).
05

Differentiate Part e

Given function is \(y = e^{5 - 6x + x^2}\). Here, \(u = 5 - 6x + x^2\), so \(\frac{du}{dx} = -6 + 2x\). The derivative \(\frac{dy}{dx} = e^{5-6x+x^2} \cdot (-6 + 2x) = e^{5-6x+x^2} \cdot (2x - 6)\).
06

Differentiate Part f

Given function is \(y = e^{\sqrt{x}}\). Here, \(u = \sqrt{x}\), so \(\frac{du}{dx} = \frac{1}{2\sqrt{x}}\). The derivative \(\frac{dy}{dx} = e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} = \frac{e^{\sqrt{x}}}{2\sqrt{x}}\).

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Key Concepts

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

Calculus
Calculus is a branch of mathematics that focuses on the study of change. At its core, calculus deals with concepts derived from limits, functions, derivatives, integrals, and infinite series.
It's divided into two major branches: differential calculus and integral calculus.
  • Differential calculus is concerned with the concept of a derivative, which measures how a function changes as its input changes. In simpler terms, it's about measuring slopes of curves or the rate of change.
  • Integral calculus, on the other hand, focuses on getting accumulated values, such as areas under and between curves.
Understanding calculus is essential for describing real-world phenomena in fields such as physics, engineering, economics, and beyond. It allows mathematicians, scientists, and engineers to find solutions to problems involving quantifiable change.
Derivatives
Derivatives are a fundamental concept in differential calculus. They represent the rate at which a function is changing at any given point, often regarded as the slope of the tangent line to the function's graph at that point.
Calculating derivatives is key to solving many problems in calculus, particularly those that involve motion and rates of change.
  • The derivative of a simple power of x, like \(x^n\), is given by \(n x^{n-1}\).
  • The exponential function \(e^x\) is unique because its derivative is simply \(e^x\) itself.
  • However, when the exponent is a function of \(x\), as seen in expressions like \(e^{3x}\), the chain rule is required to find the derivative accurately.
The computation of derivatives allows for the examination of function behaviors, optimization problems, and modeling dynamic systems.
Chain Rule
The chain rule is a powerful technique in calculus used to find the derivative of a composite function. This rule is particularly important when dealing with functions where one function is nested inside another.
The chain rule states that the derivative of the composite function \(f(g(x))\) is the derivative of \(f\) evaluated at \(g(x)\) times the derivative of \(g(x)\). Mathematically, this is expressed as:
    \[\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)\]
When working with exponential differentiation, such as finding the derivative of \(y = e^{u(x)}\), the chain rule applies as follows:
  • First, differentiate the outer function, which leaves the expression unchanged: \(e^{u(x)}\).
  • Then multiply by the derivative of the inner function, \(u'(x)\).
The chain rule provides a systematic method to differentiate complex composite functions, making it a cornerstone of differential calculus.

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Most popular questions from this chapter

Determine the equation of the tangent to the curve \(y=e^{-x}\) at the point where \(x=-1 .\) Graph the original curve and the tangent.

The squirrel population in a small self-contained forest was studied by a biologist. The biologist found that the squirrel population, \(P\), measured in hundreds, is a function of time, \(t,\) where \(t\) is measured in weeks. The function is \(P(t)=\frac{20}{1+3 e^{-0.02 t}}\) a. Determine the population at the start of the study, when \(t=0\) b. The largest population the forest can sustain is represented mathematically by the limit as \(t \rightarrow \infty .\) Determine this limit. c. Determine the point of inflection. d. Graph the function. e. Explain the meaning of the point of inflection in terms of squirrel population growth.

Differentiate each of the following functions: a. \(y=2^{3 x}\) b. \(y=3.1^{x}+x^{3}\) c. \(s=10^{3 t-5}\) d. \(w=10^{\left(5-6 n+n^{2}\right)}\) e. \(y=3^{x^{2}+2}\) f. \(y=400(2)^{x+3}\)

a. If \(f(x)=\frac{1}{3}\left(e^{3 x}+e^{-3 x}\right),\) calculate \(f^{\prime}(1)\). b. If \(f(x)=e^{-\left(\frac{1}{a+1}\right)},\) calculate \(f^{\prime}(0)\). c. If \(h(z)=z^{2}\left(1+e^{-z}\right),\) calculate \(h^{\prime}(-1)\).

Although it is true that many animal populations grow exponentially for a period of time, it must be remembered that the food available to sustain the population is limited and the population will level off because of this. Over a period of time, the population will level out to the maximum attainable value, L. One mathematical model to describe a population that grows exponentially at the beginning and then levels off to a limiting value, \(L\), is the logistic model. The equation for this model is \(P=\frac{a L}{a+(L-a) e^{-k L t}},\) where the independent variable \(t\) represents the time and \(P\) represents the size of the population. The constant \(a\) is the size of the population at \(t=0, L\) is the limiting value of the population, and \(k\) is a mathematical constant. a. Suppose that a biologist starts a cell colony with 100 cells and finds that the limiting size of the colony is 10000 cells. If the constant \(k=0.0001\) draw a graph to illustrate this population, where \(t\) is in days. b. At what point in time does the cell colony stop growing exponentially? How large is the colony at this point? c. Compare the growth rate of the colony at the end of day 3 with the growth rate at the end of day \(8 .\) Explain what is happening.
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