Problem 3
Question
Find the derivatives of the functions $$ \left(e^{x}\right)^{8} $$
Step-by-Step Solution
Verified Answer
The derivative of \((e^x)^8\) is \(8e^{8x}\).
1Step 1: Rewrite the Function
First, rewrite the function \( (e^x)^8 \) using the laws of exponents. This is equivalent to \( e^{8x} \) because \( (a^m)^n = a^{mn} \).
2Step 2: Use the Chain Rule
Differentiate the function \( e^{8x} \) using the chain rule. The chain rule states that if you have a function \( f(g(x)) \), the derivative is \( f'(g(x)) \cdot g'(x) \). Here, \( f(u) = e^u \) and \( g(x) = 8x \).
3Step 3: Differentiate the Outer Function
The derivative of \( f(u) = e^u \) with respect to \( u \) is \( e^u \). Therefore, the derivative of \( e^{8x} \) with respect to \( 8x \) is \( e^{8x} \).
4Step 4: Differentiate the Inner Function
The derivative of the inner function \( g(x) = 8x \) with respect to \( x \) is \( 8 \).
5Step 5: Combine Derivatives
Apply the chain rule by multiplying the derivative of the outer function by the derivative of the inner function: \( e^{8x} \times 8 \).
6Step 6: Simplify the Expression
Simplifying the expression gives \( 8e^{8x} \).
Key Concepts
Exponent LawsChain RuleOuter Function DerivationInner Function Derivation
Exponent Laws
Exponent laws are fundamental in simplifying expressions and finding derivatives of complex exponential functions. These laws provide us rules like \((a^m)^n = a^{mn}\), which allows us to handle powers raised to other powers.
For instance, in the given problem, the exponential expression \((e^x)^8\) is rewritten using exponent laws.
For instance, in the given problem, the exponential expression \((e^x)^8\) is rewritten using exponent laws.
- The function \((e^x)^8\) can be simplified to \(e^{8x}\), making further calculations easier.
- This transformation uses the property that the exponents multiply together: \((a^m)^n = a^{mn}\).
Chain Rule
The chain rule is an essential tool in differentiation, especially for composite functions. It provides a systematic way to differentiate functions composed of other functions.
When you have a function like \(f(g(x))\), the chain rule states that the derivative is \(f'(g(x)) \cdot g'(x)\).
When you have a function like \(f(g(x))\), the chain rule states that the derivative is \(f'(g(x)) \cdot g'(x)\).
- In our example, the function is \(e^{8x}\), represented as \(f(u)\) with \(u = 8x\).
- The chain rule helps us to multiply the derivative of the outer function by the derivative of the inner function.
Outer Function Derivation
Deriving the outer function is a key part of using the chain rule effectively. The outer function in the problem is \(f(u) = e^u\).
The derivative here is straightforward because the derivative of \(e^u\) with respect to \(u\) is simply \(e^u\).
The derivative here is straightforward because the derivative of \(e^u\) with respect to \(u\) is simply \(e^u\).
- This simplicity makes exponential functions like \(e^u\) very convenient to differentiate.
- Thus, for our problem, we find that the derivative of \(e^{8x}\) with respect to its inner component is \(e^{8x}\).
Inner Function Derivation
To complete our use of the chain rule, we need to differentiate the inner function. Here, the inner function is \(g(x) = 8x\).
Deriving this function is simple, as it is a linear expression. The derivative of \(8x\) with respect to \(x\) is just \(8\).
Deriving this function is simple, as it is a linear expression. The derivative of \(8x\) with respect to \(x\) is just \(8\).
- This step provides the scaling factor needed in the chain rule multiplication.
- Understanding inner function derivation aids in logically reconstructing complex function derivatives.
Other exercises in this chapter
Problem 3
Solve staring from \(y_{0}=1\) and from \(y_{0}=-1 .\) Draw both solution on the same graph. $$ \frac{d y}{d t}=2 y $$
View solution Problem 3
Find the derivative \(d y / d x\) in \(1-10\). $$ y=(\ln x)^{-1} $$
View solution Problem 3
Sketch \(y=2^{-x}\) and \(y=\frac{1}{2}\left(4^{x}\right)\) from -1 to 1 on the same graph. Put their mirror images \(x=-\log _{2} y\) and \(x=\log _{4} 2 y\) o
View solution Problem 4
By the quotient rule, verify \((\tanh x)^{\prime}=\operatorname{sech}^{2} x\).
View solution