/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 85 Differentiate. $$ y=\sin \le... [FREE SOLUTION] | 91Ó°ÊÓ

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Differentiate. $$ y=\sin \left(\cos e^{x}\right) $$

Short Answer

Expert verified
\( \frac{dy}{dx} = -\cos(\cos(e^{x})) \cdot \sin(e^{x}) \cdot e^{x}\)

Step by step solution

01

Identify the outer function

The given function is a composite function. Identify the outermost function. Here, the outer function is the sine function, \( \sin(u) \).
02

Differentiate the outer function

Differentiate \( \sin(u) \) with respect to \( u \): \( \frac{d}{du} \sin(u) = \cos(u)\. \) Here, \( u = \cos(e^{x}) \).
03

Identify the inner function

Next, identify the inner function inside the sine function, which is \( \cos(e^{x}) \).
04

Differentiate the inner function

Differentiate the inner function \( \cos(e^{x}) \) with respect to \( x \): \( \frac{d}{dx} \cos(e^{x}) = -\sin(e^{x}) \cdot \frac{d}{dx}(e^{x}) = -\sin(e^{x}) \cdot e^{x} \).
05

Use the chain rule

Apply the chain rule to combine the derivatives from steps 2 and 4. The chain rule states that \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \). Therefore, \ \frac{dy}{dx} = \cos(\cos(e^{x})) \cdot [-\sin(e^{x}) \cdot e^{x}]\.
06

Simplify the expression

Finally, simplify the expression: \[ \frac{dy}{dx} = -\cos(\ \cos(e^{x}) \ ) \cdot \sin(e^{x}) \cdot e^{x} \]\.

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Key Concepts

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

Chain Rule
Understanding the Chain Rule is central to differentiating composite functions. A composite function is one where a function is applied to another function, like our given function, \( y = \sin(\cos(e^{x})) \). The Chain Rule states that to differentiate a composite function, you multiply the derivative of the outer function by the derivative of the inner function. Mathematically, if you have a function \( y = f(g(x)) \), you need to find \( f'(g(x)) \) and multiply by \( g'(x) \). This allows you to break down complex derivatives into simpler steps, which you can then combine for the final result.
Calculus for Life Sciences
In life sciences, understanding how quantities change with respect to each other is key. Differentiation allows for precise modeling of biological phenomena, such as growth rates, enzyme kinetics, and population dynamics. For example, knowing how the concentration of a substance changes with time can help in developing effective drug dosages. Differentiating composite functions could be useful to determine how nested biological processes interact and change over time. When applying calculus in life sciences, always aim to understand the relationship between the variables you're differentiating.
Derivative of Exponential Functions
Exponential functions, such as \( e^{x} \), frequently appear in various scientific fields. The unique characteristic of the exponential function \( e^{x} \) is that it’s its own derivative, meaning \( \frac{d}{dx} e^{x} = e^{x} \). When you have an exponential function nested within another function, like in our example, you also need to differentiate it as part of the Chain Rule. For instance, \( \frac{d}{dx} \cos(e^{x}) \), requires differentiating \( e^{x} \) first, then multiplying by the derivative of the cosine function to get \( -\sin(e^{x}) \cdot e^{x} \). Understanding this helps you handle more complex differentiation problems.
Trigonometric Function Derivatives
Trigonometric functions, such as sine and cosine, have specific rules for differentiation. The derivative of the sine function, \( \sin(x) \), is the cosine function, \( \cos(x) \). Meanwhile, the derivative of the cosine function, \( \cos(x) \), is \( -\sin(x) \). When these trigonometric functions become part of a composite function, you must apply the Chain Rule. In our example, differentiating \( \sin(\cos(e^{x})) \), involved using these trigonometric derivatives along with exponential function derivatives, resulting in the final expression: \( \frac{dy}{dx} = -\cos(\cos(e^{x})) \cdot \sin(e^{x}) \cdot e^{x} \). This highlights how combining these principles provides a clear pathway to the solution.

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Most popular questions from this chapter

The growth of maize (corn) leaves is modeled by the logistic equation. For the sixth leaf on a maize plant, it was found that the function $$ A(t)=\frac{105}{1+32900 e^{-0.04 t}} $$ is a good approximation of the leaf's area \(A(t)\) (in \(\mathrm{cm}^{2}\) ) from the time the leal forms until it stops growing. \({ }^{23}\) The number \(t\) is the number of degree days since the plant sprouted. (Degree days measure time weighted by the temperature. The number of degree days for maize is computed by multiplying the time by the number of degrees above \(8^{\circ} \mathrm{C}\).) a) Use the logistic equation to approximate the leaf area after 100 degree days, after 150 degree days, and after 200 degree days. b) Find the rate of change of the area \(A^{\prime}(t)\). c) Find the point of inflection for the function \(A(t)\) d) Sketch the graph of \(A(t)\). e) According to this model, what is the maximum area of the sixth leal on a maize plant?

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