Problem 5 Compute \(y^{\prime}(x)\) or ass... [FREE SOLUTION]

Chapter 5: Problem 5

Compute \(y^{\prime}(x)\) or assert that you do not yet have forumlas to compute \(\mathbf{y}^{\prime}(\mathbf{x})\) for a. \(y(x)=e^{5 x}\) b. \(y(x)=e^{-3 x}\) c. \(y(x)=e^{\sqrt{x}}\) d. \(y(x)=\left(e^{x}\right)^{2}\) e. \(y(x)=\left(e^{\sqrt{x}}\right)^{2}\) f. \(y(x)=\left(e^{-x}\right)^{2}\) g. \(y(x)=\frac{e^{x}+e^{-x}}{2}\) h. \(y(x)=\frac{e^{x}-e^{-x}}{2}\) i. \(y(x)=5 e^{-0.06 x}+3 e^{-0.1 x} \quad\) j. \(y(x)=e^{\left(x^{2}\right)}\) k. \(y(x)=\sqrt{e^{x}}\) l. \(y(x)=8 e^{-0.0001 x}-16 e^{-0.001 x}\) m. \(y(x)=e^{5}\) n. \(y(x)=\sqrt{e}\) o. \(y(x)=10^{x}\) p. \(y(x)=10^{-x}\) q. \(y(x)=x^{2}+2^{x}\) r. \(y(x)=\left(e^{5 x}+e^{-3 x}\right)^{5}\)

Short Answer

Expert verified

Differentiate using the chain rule and known derivatives: 1. \(5e^{5x}\), 2. \(-3e^{-3x}\), 3. \(e^{\sqrt{x}} \cdot \frac{1}{2}x^{-1/2}\), ... 16. \(5(e^{5x}+e^{-3x})^4(5e^{5x} - 3e^{-3x})\).

Step by step solution

Differentiate e^{5x}

For function a, use the chain rule. The derivative of \(e^{u}\) is \(e^{u} \cdot u^{\prime}\) where \(u = 5x\). We have \(u^{\prime} = 5\). Thus, \(y^{\prime}(x) = 5e^{5x}\).

Differentiate e^{-3x}

For function b, again use the chain rule. The derivative of \(e^{u}\) is \(e^{u} \cdot u^{\prime}\) where \(u = -3x\). We have \(u^{\prime} = -3\). Thus, \(y^{\prime}(x) = -3e^{-3x}\).

Differentiate e^{\sqrt{x}}

For function c, use the chain rule. Here, \(u = \sqrt{x} = x^{1/2}\) and \(u^{\prime} = \frac{1}{2}x^{-1/2}\). Thus, \(y^{\prime}(x) = e^{\sqrt{x}} \cdot \frac{1}{2}x^{-1/2}\).

Differentiate (e^x)^2

For function d, \((e^{x})^{2} = e^{2x}\). Differentiate as earlier: \(u = 2x\), \(u^{\prime} = 2\). Thus, \(y^{\prime}(x) = 2e^{2x}\).

Differentiate (e^{\sqrt{x}})^2

For function e, use the chain rule. Let \(u = e^{\sqrt{x}}\), then \(y = u^{2}\) and \(y^{\prime}(x) = 2u \cdot u^{\prime}\). Since \(u^{\prime} = e^{\sqrt{x}} \cdot \frac{1}{2}x^{-1/2}\), substitute to get \(y^{\prime}(x) = 2e^{\sqrt{x}} \cdot e^{\sqrt{x}} \cdot \frac{1}{2}x^{-1/2} = e^{2\sqrt{x}}x^{-1/2}\).

Differentiate (e^{-x})^2

For function f, \((e^{-x})^{2} = e^{-2x}\). Differentiate using \(u = -2x\) and \(u^{\prime} = -2\). Thus, \(y^{\prime}(x) = -2e^{-2x}\).

Differentiate \frac{e^{x} + e^{-x}}{2}

For function g, differentiate each part: \(\frac{d}{dx}e^{x} = e^{x}\) and \(\frac{d}{dx}e^{-x} = -e^{-x}\). The derivative is \(\frac{1}{2}(e^{x} - e^{-x})\).

Differentiate \frac{e^{x} - e^{-x}}{2}

For function h, similarly to g, \(\frac{1}{2}(e^{x} + e^{-x})\).

Differentiate 5e^{-0.06x} + 3e^{-0.1x}

For function i, use the linearity of differentiation. \(\frac{d}{dx}(5e^{-0.06x}) = -0.3e^{-0.06x}\) and \(\frac{d}{dx}(3e^{-0.1x}) = -0.3e^{-0.1x}\). Add them up to get \(-0.3e^{-0.06x} - 0.3e^{-0.1x}\).

Differentiate e^{x^2}

For function j, use the chain rule. Here, \(u = x^{2}\) and \(u^{\prime} = 2x\). Thus, \(y^{\prime}(x) = 2xe^{x^{2}}\).

Differentiate \sqrt{e^{x}}

For function k, \(y = e^{x/2}\) (since \(\sqrt{a} = a^{1/2}\)). Differentiate: \(u = x/2\), \(u^{\prime} = 1/2\). \(y^{\prime}(x) = \frac{1}{2}e^{x/2}\).

Differentiate 8e^{-0.0001x} - 16e^{-0.001x}

For function l, differentiate each term: \(\frac{d}{dx}(8e^{-0.0001x}) = -0.0008e^{-0.0001x}\) and \(\frac{d}{dx}(-16e^{-0.001x}) = 0.016e^{-0.001x}\). Combine to get \(-0.0008e^{-0.0001x} + 0.016e^{-0.001x}\).

Differentiate constants e^{5} and \sqrt{e}

For m and n, they are constants. The derivative of a constant is 0.

Differentiate 10^x and 10^{-x}

For function o, use \(y = a^{x}\) where \(a = 10\), giving \(y^{\prime} = 10^{x} \ln(10)\). For function p, \(y = 10^{-x}\), yielding \(y^{\prime} = -10^{-x} \ln(10)\).

Differentiate x^2 + 2^x

For function q, differentiate each term individually: \(y^{\prime}(x) = 2x + 2^{x} \ln(2)\).

Differentiate (e^{5x} + e^{-3x})^5

For function r, use the chain rule. Let \(u = e^{5x} + e^{-3x}\), then \(y = u^{5}\) and \(y^{\prime} = 5u^{4} \cdot (5e^{5x} - 3e^{-3x})\).

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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 fundamental tool in calculus, allowing us to differentiate composite functions. A composite function is simply a function within another function. The main idea is to differentiate the outer function and multiply it by the derivative of the inner function. For example, if we have a function like \(y = e^{5x}\), we treat \(5x\) as the inner function and \(e^u\) as the outer function, where \(u = 5x\). Using the chain rule, the derivative is \(e^{5x} imes 5\), which simplifies to \(5e^{5x}\). This rule helps break down the differentiation process into manageable steps and is particularly useful when dealing with exponential functions.

Exponential Functions

Exponential functions have the form \(y = a^x\), where \(a\) is a positive real number not equal to 1. These functions have a unique property: their rate of growth or decay is directly proportional to their current value. In mathematics, this is often expressed through the natural exponential function, \(e^x\). The base \(e\) is an irrational constant approximately equal to 2.71828. When differentiating exponential functions, such as \(e^{u}\), the derivative is \(e^{u} imes u'\). For example, in the function \(y = e^{x^2}\), the derivative would be found by treating \(x^2\) as the inner function, giving us \(2x imes e^{x^2}\). The exponential function's behavior in calculus makes it both predictable and widely applicable.

Derivative of a Constant

Finding the derivative of a constant is one of the simplest tasks in calculus. A constant is a fixed value that does not change, and its graph is a horizontal line. Because the slope of a horizontal line is zero, the derivative of any constant is always zero. For instance, if you have a constant function \(y = e^5\) or \(y = \sqrt{e}\), regardless of the value of \(e\), its derivative remains zero: \(y' = 0\). This concept is important as it simplifies the differentiation process, allowing us to focus on the variable components of more complex expressions.

Power Rule

The Power Rule is another critical component of differentiation. It simplifies the process of finding the derivative of a power function. If you have a function in the form of \(y = x^n\), where \(n\) is a real number, the derivative is obtained by multiplying \(n\) by the power \(x\) raised to the power of \(n-1\). In mathematical terms, \(y' = nx^{n-1}\). For example, if \(y = x^2\), applying the power rule gives \((y' = 2x^{2-1} = 2x)\). This rule extends beyond simple polynomials and can be applied to functions where the variable base has non-integer powers, like \(y = \sqrt{x}\) or \(y = x^{1/2}\). The power rule is a powerful tool, making it easy to handle a wide range of algebraic expressions.

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