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Find the derivatives of the functions. Assume that \(a, b, c,\) and \(k\) are constants. $$y(x)=a^{x}+x^{a}$$

Short Answer

Expert verified
The derivative is \(y'(x) = a^{x}\ln(a) + a \cdot x^{a-1}\).

Step by step solution

01

Differentiate the first term

For the function \(y(x) = a^{x} + x^{a}\), we first differentiate the term \(a^{x}\). The derivative of \(a^{x}\) with respect to \(x\) is \(a^{x} \ln(a)\). Use the formula for differentiating exponential functions \(\frac{d}{dx}[a^x] = a^x \ln(a)\).
02

Differentiate the second term

Now, differentiate the term \(x^{a}\). This is a power function where \(a\) is treated as a constant. Apply the power rule: \(\frac{d}{dx}[x^a] = a \cdot x^{a-1}\).
03

Combine the derivatives

Add the derivatives from Step 1 and Step 2 to find the derivative of the entire function. Therefore, the derivative \(y'(x)\) is given by:\[ y'(x) = a^{x} \ln(a) + a \cdot x^{a-1}. \]

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Key Concepts

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

Exponential Functions
Exponential functions are essential topics in calculus and appear in different mathematical contexts. An exponential function has the general form \(f(x) = a^x\), where \(a\) is a constant. These functions have the unique characteristic of growing rapidly due to the exponent variable being in the power position. When we take the derivative of an exponential function like \(a^x\), we utilize a specific rule involving the natural logarithm. The derivative of \(a^x\) is \(a^x \ln(a)\). Understanding the role of \( \ln(a) \) is crucial. It stems from the change in rate of \(a^x\) as \(x\) increases. This rule is a foundation in calculus, making it possible to analyze the behavior of exponential growth in multiple applications, from population growth models to finance.
To summarize:
  • Exponential functions take the form \(a^x\).
  • The derivative of an exponential function is \(a^x \ln(a)\).
Power Functions
Power functions are described by expressions like \(x^a\), where \(a\) is a real number. These functions are versatile, covering a wide range of problems in mathematics.
To find the derivative of power functions, we apply the power rule of differentiation. This rule states that \[ \frac{d}{dx}[x^a] = a \cdot x^{a-1}. \] This result occurs because each time we differentiate, we decrease the exponent by one and multiply by the original exponent. Power functions and their derivatives are foundational in calculus, influencing calculations in physics, engineering, and more. Knowing how to differentiate these functions efficiently can reveal many real-world phenomena and mathematical insights.
In summary:
  • Power functions are in the form \(x^a\).
  • The power rule gives us \( \frac{d}{dx}[x^a] = a \cdot x^{a-1}\).
Differentiation Rules
Differentiation in calculus is about calculating rates of change. It involves using specific rules to find derivatives, which represent how a function changes instantly at any point. Two primary rules we've used are the exponential function rule and the power rule.
The exponential rule for differentiation, employed with functions like \(a^x\), depends on the function being in an exponent form, leading to the derivative \(a^x \ln(a)\). On the other hand, the power rule applies to power functions like \(x^a\), where derivatives are found using \(a \cdot x^{a-1}\).
Combining these rules allows us to differentiate more complex expressions, like \(y(x) = a^x + x^a\), by applying the rules to each term separately. Finally, we sum these individual derivatives to find the overall function's derivative. These techniques are invaluable, whether you're solving textbook problems or analyzing scientific data.
Key differentiation rules include:
  • Exponential rule: \( \frac{d}{dx}[a^x] = a^x \ln(a) \)
  • Power rule: \( \frac{d}{dx}[x^a] = a \cdot x^{a-1} \)
  • Sum of derivatives rule for combining derivative results from each term.

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