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Find the derivative of the transcendental function. $$ f(x)=-e^{x}+\tan x $$

Short Answer

Expert verified
The derivative of the function is \(f'(x)=-e^{x}+\sec^2 x\).

Step by step solution

01

Identify component functions

The given function can be written as \(f(x)=-e^{x}+\tan x\). Here, we have two component functions: \(f_1(x)=-e^{x}\) and \(f_2(x)=\tan x\)
02

Differentiate exponential function

The derivative of \(f_1(x)=-e^{x}\) is simply \(-e^{x}\), since the derivative of \(e^{x}\) is itself \(e^{x}\). Hence, the derivative of \(f_1(x)=-e^{x}\) is \(-e^{x}\).
03

Differentiate trigonometric function

The derivative of \(f_2(x)=\tan x\) is \(\sec^2 x\), according to the standard derivative rules for the trigonometric function.
04

Writing the derivative of the combined function

The derivative of the entire function \(f(x)=-e^{x}+\tan x\) is the sum of the derivatives of the individual components. Therefore, deriving \(f(x)\) yields \(-e^{x}+\sec^2 x\).

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Key Concepts

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

Differentiate Exponential Function
Understanding how to differentiate the exponential function is essential in calculus, especially when dealing with transcendental functions. The exponential function, often represented as \( e^{x} \), has a unique property: its derivative is the same as the original function. So, when you differentiate \( e^{x} \) with respect to x, the result is simply \( e^{x} \).

In the context of our example, where the exponential function is multiplied by -1, yielding \( f_1(x) = -e^{x} \), the differentiation process remains straightforward. Applying the constant multiplier rule, which states that the derivative of a constant multiplied by a function is the constant multiplied by the derivative of the function, yields the derivative \( -e^{x} \). This rule allows us to easily manage exponential functions even when they are scaled by coefficients.
Differentiate Trigonometric Function
Trigonometric functions, which are relations of angles and sides of triangles, also play a pivotal role in calculus. Specifically, the tangent function, denoted as \( \tan(x) \), has its own differentiation rule.

To find the derivative of \( \tan(x) \), you can rely on its established derivative rule, which states that the derivative of \( \tan(x) \) is \( \sec^2(x) \). This rule is derived from the quotient rule because \( \tan(x) \) can also be expressed as the ratio of \( \sin(x) \) over \( \cos(x) \). Considering that trigonometric functions can often result in more complex derivatives due to their periodic nature, recognizing and applying these derivative rules is key to simplifying calculus problems.
Calculus Problems and Solutions
When solving calculus problems involving derivatives, it's important to apply differentiation rules systematically. To solve for the derivative of a transcendental function like \( f(x)=-e^{x}+\tan x \), as seen in our example, you must differentiate each component function separately and then combine their derivatives appropriately.

This process usually involves identifying the type of each component (exponential, trigonometric, etc.), applying the correct differentiation rules for each type, and then using the sum rule to add or subtract the resulting derivatives, depending on how the components combine in the original function. The sum rule is a principle that states that the derivative of a sum is the sum of the derivatives. Following a step-by-step approach ensures that you have a clear pathway to find the derivative, helping you avoid mistakes and better understand the calculus behind the problem.

Remember, practicing a variety of problems with different functions is the best way to become proficient at solving calculus equations. This hands-on experience reinforces your understanding of the differentiation rules and how they apply to various functions.

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Most popular questions from this chapter

True or False? In Exercises 137-139, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If \(y\) is a differentiable function of \(u, u\) is a differentiable function of \(v,\) and \(v\) is a differentiable function of \(x,\) then \(\frac{d y}{d x}=\frac{d y}{d u} \frac{d u}{d v} \frac{d v}{d x}\)

Linear and Quadratic Approximations The linear and quadratic approximations of a function \(f\) at \(x=a\) are \(P_{1}(x)=f^{\prime}(a)(x-a)+f(a)\) and \(P_{2}(x)=\frac{1}{2} f^{\prime \prime}(a)(x-a)^{2}+f^{\prime}(a)(x-a)+f(a)\) \(\begin{array}{llll}\text { In Exercises } & 133-136, & \text { (a) find the specified linear and }\end{array}\) quadratic approximations of \(f,\) (b) use a graphing utility to graph \(f\) and the approximations, (c) determine whether \(P_{1}\) or \(P_{2}\) is the better approximation, and (d) state how the accuracy changes as you move farther from \(x=a\). $$ \begin{array}{l} f(x)=\sec 2 x \\ a=\frac{\pi}{6} \end{array} $$

In Exercises \(81-88\), (a) find an equation of the tangent line to the graph of \(f\) at the indicated point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of a graphing utility to confirm your results. \(\frac{\text { Function }}{f(x)=\sqrt{3 x^{2}-2}} \quad \frac{\text { Point }}{(3,5)}\)

The area of a square with sides of length \(s\) is given by \(A=s^{2} .\) Find the rate of change of the area with respect to \(s\) when \(s=4\) meters.

This law states that the rate of change of the temperature of an object is proportional to the difference between the object's temperature \(T\) and the temperature \(T_{a}\) of the surrounding medium. Write an equation for this law.

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