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In Exercises find \(d y / d x\). $$y=e^{x \tan x}$$

Short Answer

Expert verified
\( \frac{dy}{dx} = e^{x \tan x} (\tan x + x \sec^2 x) \)

Step by step solution

01

Apply the Chain Rule

The given function is a composite function where the outer function is an exponential function and the inner function is the product of the variables inside the exponent. Recognize that to differentiate \( y = e^{x \tan x} \), you need to use the chain rule. The chain rule states that \( \frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x) \). Here, \( f(u) = e^u \) and \( u = x \tan x \).
02

Differentiate the Outer Function

Differentiate the outer function \( e^u \) with respect to \( u \). The derivative of \( e^u \) is itself, so \( \frac{d}{du} [e^u] = e^u \).
03

Differentiate the Inner Function

Differentiate the inner function \( u = x \tan x \) with respect to \( x \). To do this, you need to apply the product rule: \( \frac{d}{dx}[uv] = u'v + uv' \). Here, \( u = x \) and \( v = \tan x \). Calculate \( \frac{d}{dx}(x) = 1 \) and \( \frac{d}{dx}(\tan x) = \sec^2 x \). The derivative of \( x \tan x \) is \( 1 \cdot \tan x + x \cdot \sec^2 x \). This simplifies to \( \tan x + x \sec^2 x \).
04

Combine the Derivatives

Now use the results from the chain rule: \( \frac{dy}{dx} = e^{x \tan x} \cdot (\tan x + x \sec^2 x) \). Replace \( u \) back with \( x \tan x \) and combine: \( \frac{dy}{dx} = e^{x \tan x} \cdot (\tan x + x \sec^2 x) \).
05

Simplify the Expression

Finally, write down the simplified version of the derivative, \( \frac{dy}{dx} = e^{x \tan x} (\tan x + x \sec^2 x) \), which cannot be simplified further.

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Key Concepts

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

Differentiation
Differentiation is a fundamental concept in calculus that involves finding the rate at which a function changes. It's like finding the slope of a line at any given point on a curve. In a more practical sense, it allows us to see how changes in one quantity affect another. Differentiation is often used in various fields, such as physics, engineering, and economics, to model and analyze dynamic systems.

When differentiating a function, we look for its derivative, denoted typically by \(\frac{dy}{dx}\), which tells us how \(y\) changes as \(x\) changes. The process of finding a derivative involves applying specific rules, such as the power rule, product rule, chain rule, and others, depending on the form of the function being differentiated. By understanding differentiation, one can solve problems involving rates of change efficiently.
Product Rule
The product rule is a helpful differentiation technique when dealing with functions that are products of two or more simpler functions. If a function can be expressed as \(uv\), where both \(u\) and \(v\) are functions of \(x\), the product rule states that the derivative is given by:
  • \(\frac{d}{dx}[uv] = u'v + uv'\)
Here, \(u'\) represents the derivative of \(u\) with respect to \(x\), and \(v'\) represents the derivative of \(v\) with respect to \(x\).

Applying the product rule ensures that you account for how both functions \(u\) and \(v\) change. In practice, this means you take the derivative of each function separately, multiply it by the other function, and then add the results together. This rule is crucial when the terms being multiplied do not need further simplification to differentiate, as seen in problems like finding \(\frac{dy}{dx}\) for \(x \tan x\).
Exponential Function
Exponential functions are a key part of calculus, often characterized by swift growth or decay. Their standard form is \(e^x\), where \(e\) is Math's natural constant, approximately equal to 2.718. These functions are particularly interesting because their derivatives are incredibly neat: the derivative of an exponential function with respect to \(x\) is the function itself. So, when we have \(y = e^x\), its derivative \(\frac{d}{dx}[e^x]\) is simply \(e^x\).

In more complex functions where the exponent is itself a function of \(x\), such as \(e^{g(x)}\), we must use the chain rule to find the derivative. This involves differentiating the exponent separately, then multiplying by the derivative of the exponential function. Exponential functions are commonly used in modeling population growth, radioactive decay, and compound interest.
Composite Function
A composite function arises when one function is nested inside another. It's a bit like peeling an onion, where you deal with each layer separately. In calculus, differentiating composite functions requires the chain rule, a helpful method for untangling these nested functions.

To simplify, imagine a function \(f(g(x))\), where \(f\) is applied to \(g(x)\). The chain rule states that the derivative is:
  • \(\frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x)\)
What this means is that you first find the derivative of the outer function \(f\), then multiply it by the derivative of the inner function \(g\).

For example, in \(y = e^{x \tan x}\), "\(e^u\)" is the outer function, and "\(x \tan x\)" is the inner. Recognizing and appropriately handling composite functions like these can make complex problems far more approachable.

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Most popular questions from this chapter

Two curves are said to be orthogonal if their tangent lines are perpendicular at each point of intersection, and two families of curves are said to be orthogonal trajectories of one another if each member of one family is orthogonal to each member of the other family. The accompanying figure shows some typical members of the families of hyperbolas \(x y=c\) (black curves) and \(x^{2}-y^{2}=k\) (gray curves), where \(c \neq 0\) and \(k \neq 0 .\) Use the hint in Exercise 51 to show that these families are orthogonal trajectories of one another. GRAPH CANT COPY

A point \(P\) is moving along the line whose equation is \(y=2 x .\) How fast is the distance between \(P\) and the point (3.0) changing at the instant when \(P\) is at (3,6) if \(x\) is decreasing at the rate of 2 units/s at that instant?

Find the limit. $$\lim _{x \rightarrow+\infty}[\ln x-\ln (1+x)]$$

The equation \(Q=12 e^{-0.055 t}\) gives the mass \(Q\) in grams of radioactive potassium- 42 that will remain from some initial quantity after \(t\) hours of radioactive decay. (a) How many grams were there initially? (b) How many grams remain after 4 hours? (c) How long will it take to reduce the amount of radioactive potassium- 42 to half of the initial amount?

There is a myth that circulates among beginning calculus students which states that all indeterminate forms of types \(0^{0}, \infty^{0},\) and \(1^{*}\) have value 1 because "anything to the zero power is \(1^{\prime \prime}\) and \(" 1\) to any power is \(1 . "\) The fallacy is that \(0^{0}, \infty^{0},\) and \(1^{\times x}\) are not powers of numbers, but rather descriptions of limits. The following examples, which were transmitted to me by Prof. Jack Staib of Drexel University, show that such indeterminate forms can have any positive real value: (a) \(\lim _{x \rightarrow 0^{+}}\left[x^{(\ln a) /(1+\ln x)}\right]=0^{0}=a\) (b) \(\lim _{x \rightarrow+\infty}\left[x^{(\ln a) /(1+\ln x)}\right]=\infty^{0}=a\) (c) \(\lim _{x \rightarrow 0}\left[(x+1)^{(\ln a) / x}\right]=1^{x}=a\) Prove these results.

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