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Find the derivative of each function. \(e^{x^{3}+6 x}\)

Short Answer

Expert verified
The derivative is \( f'(x) = e^{x^3 + 6x} \cdot (3x^2 + 6) \).

Step by step solution

01

Identify the Function Form

The given function is \( f(x) = e^{x^3 + 6x} \). This is an exponential function with a composite exponent, \( x^3 + 6x \).
02

Apply the Chain Rule

To differentiate \( f(x) = e^{x^3 + 6x} \), use the chain rule: If \( g(x) = e^{u(x)} \), then \( g'(x) = e^{u(x)} \cdot u'(x) \). Here, \( u(x) = x^3 + 6x \).
03

Differentiate the Inner Function

Find the derivative of the inner function \( u(x) = x^3 + 6x \). Using basic differentiation rules, \( u'(x) = 3x^2 + 6 \).
04

Combine Results to Find the Derivative

Substitute \( u(x) \) and \( u'(x) \) back into the chain rule result: \( f'(x) = e^{x^3 + 6x} \cdot (3x^2 + 6) \). This gives the derivative of the original function.

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Key Concepts

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

Differentiation Essentials
Differentiation is one of the core concepts in calculus, focused on computing the derivative of a function. A derivative measures how a function's output value changes as its input value changes. This is crucial for understanding the behavior of functions, such as determining slopes of curves and rates of change.

When differentiating, you're essentially finding the rate at which the function's output is changing for every point on its curve. Differentiation can be thought of as a mathematical tool that answers the question "How fast is this quantity changing?" or "What is the slope of the tangent line at a particular point on a curve?".

To differentiate a function, apply specific rules depending on the type of function—these include power rules, constant rules, sum rules, and more. However, there's a special rule, the chain rule, which can be particularly handy for functions made up of multiple sub-functions, called composite functions.
Understanding the Chain Rule
The chain rule is essential for differentiating composite functions—these are functions where one function is inside another. Imagine it as a chain where each link is a function. When you differentiate such a function, the chain rule helps you "unravel" it, differentiating each link step by step.

The chain rule states that if you have a function that can be written as a composition of two functions, say \[y = f(g(x))\], then the derivative can be determined by \[dy/dx = f'(g(x)) \cdot g'(x)\].

Think of it like peeling an onion: you differentiate from the outside inwards. You first differentiate the outer function, keeping the inner function unchanged, and then multiply the result by the derivative of the inner function. This technique is especially useful when dealing with complex functions, ensuring each component's influence is accurately captured in the derivative.
Exploring Exponential Functions
Exponential functions are a special class of functions in mathematics that feature prominently in calculus. These functions have the form \[f(x) = a^{g(x)}\], where 'a' is a constant, and the exponent \( g(x) \) is a function of 'x'. The most common base used is the mathematical constant 'e', approximately equal to 2.718.

In the world of calculus, exponential functions are fascinating because their rate of growth is proportional to their current value. This makes them unique for modeling real-world situations involving constant relative growth rates, such as populations or compound interest.

When differentiating an exponential function of the form \[e^{g(x)}\], it combines with the chain rule. Start by noting that the derivative of \[e^{x}\] is simply \[e^{x}\] itself. So, \[f'(x) = e^{g(x)} \cdot g'(x)\]. This intertwined relationship of the exponential function and the chain rule often leads to efficient and insightful solutions in calculus problems.

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Most popular questions from this chapter

The substitution method can be used to find integrals that do not fit our formulas. For example, observe how we find the following integral using the substitution \(u=x+4\) which implies that \(x=u-4\) and so \(d x=d u\). $$ \begin{aligned} \int(x-2)(x+4)^{8} d x &=\int(u-4-2) u^{8} d u \\ &=\int(u-6) u^{8} d u \\ &=\int\left(u^{9}-6 u^{8}\right) d u \\ &=\frac{1}{10} u^{10}-\frac{2}{3} u^{9}+C \\ &=\frac{1}{10}(x+4)^{10}-\frac{2}{3}(x+4)^{9}+C \end{aligned} $$ It is often best to choose \(u\) to be the quantity that is raised to a power. The following integrals may be found as explained on the left (as well as by the methods of Section 6.1). $$ \int(x-1) \sqrt{x+2} d x $$

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