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Use Version I of the Chain Rule to calculate \(\frac{d y}{d x}\). $$y=\sin ^{5} x$$

Short Answer

Expert verified
Answer: The derivative of y with respect to x, where y = sin^5(x) is dy/dx = 5sin^4(x)cos(x).

Step by step solution

01

Identify the inner and outer functions

Let's rewrite the given function: y = sin^5(x) as the composition of two functions. We can think of the function as y = (sin(x))^5. Here, the inner function, u, is sin(x) and the outer function, v, is v = u^5.
02

Find the derivatives of the inner and outer functions

Now, we need to find the derivatives of both u and v with respect to x. For the inner function u=sin(x), the derivative is given by \(\frac{d u}{d x} = \cos(x)\). For the outer function v=u^5, the derivative is given by \(\frac{d v}{d u} = 5u^4\).
03

Apply the Chain Rule

Now, we apply the Chain Rule to find \(\frac{d y}{d x} = \frac{d v}{d u} \cdot \frac{d u}{d x}\). Plug in the derivatives we found in Step 2: \(\frac{d y}{d x} = 5u^4 \cdot \cos(x)\).
04

Replace u with the inner function

Lastly, we need to replace u back with the original inner function sin(x): \(\frac{d y}{d x} = 5(\sin(x))^4 \cdot \cos(x)\). Thus, the derivative of \(y=\sin^5(x)\) with respect to x is: $$\frac{d y}{d x} = 5\sin^4(x) \cos(x)$$

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Key Concepts

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

Derivative of Composite Functions
Understanding the derivative of composite functions is essential in calculus, as it involves the differentiation of functions that are nested within one another. This process is handled by the Chain Rule, an indispensable tool for calculus students. The Chain Rule dictates that if a function y is composed of a function of another function, such as y = f(g(x)), then the derivative of y with respect to x is the product of the derivative of the outer function with respect to the inner function and the derivative of the inner function with respect to x, formally expressed as \( \frac{dy}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx} \). This method is vitally important for tackling complex differential calculus problems where simple differentiation rules don't apply directly. In our exercise, the function \( y=\sin^5(x) \) is viewed as \( f(g(x)) \), with \( g(x)=\sin(x) \) as the inner function and \( f(u)=u^5 \) as the outer function, where \( u=g(x) \). The calculated derivative, \( 5\sin^4(x) \cos(x) \), exemplifies how the Chain Rule effectively decomposes the process into more manageable steps.
Calculus Differentiation Techniques
Calculus differentiation techniques encompass more than just the Chain Rule. There is an array of strategies to differentiate various kinds of functions. Some common techniques include the Power Rule, Product Rule, Quotient Rule, and the derivatives of inverse, exponential, logarithmic, and trigonometric functions.Each technique is tailored for different scenarios, such as using the Product Rule when differentiating the product of two functions, or the Quotient Rule for their quotient. When approaching differentiation problems, the choice of technique is pivotal for simplifying the process and ensuring accuracy. For instance, the Power Rule quickly calculates derivatives of power functions, while the Chain Rule is a go-to method for composite functions. Mastery of these techniques is beneficial for a deeper understanding and for tackling more advanced calculus problems.
Trigonometric Functions Derivatives
The derivatives of trigonometric functions are specific rules used in calculus to calculate the rates of change of the sine, cosine, tangent, and other trigonometric functions. The basic derivatives that every student should know include:
  • \( \frac{d}{dx}(\sin x) = \cos x \)
  • \( \frac{d}{dx}(\cos x) = -\sin x \)
  • \( \frac{d}{dx}(\tan x) = \sec^2 x \)
These fundamental derivatives can be combined with other differentiation techniques, like the Chain Rule, to tackle more complicated functions involving trigonometry. In our exercise, we specifically used the derivative of the sine function to find the derivative of \( y=\sin^5(x) \). Recognizing and applying these derivatives is crucial in various fields, such as physics and engineering, where trigonometric functions often describe wave patterns and oscillations.
Power Rule
The Power Rule is one of the most fundamental differentiation rules in calculus. It states that for any real number n, the derivative of \( x^n \) with respect to x is \( nx^{n-1} \). Simply put, to differentiate \( x^n \), you multiply the expression by the exponent and then subtract one from that exponent. It's concise, efficient, and applies to both positive and negative exponents provided that the base is a variable raised to a constant power.The Power Rule is extremely helpful for polynomial functions and also plays a vital role whenever an expression within a function can be expressed as a power function. For example, in our textbook problem, the outer function \( v=u^5 \) was differentiated using the Power Rule, resulting in \( 5u^4 \). Combining this with other differentiation techniques such as the Chain Rule facilitates the differentiation of even more complex functions.

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Most popular questions from this chapter

Identifying functions from an equation. The following equations implicitly define one or more functions. a. Find \(\frac{d y}{d x}\) using implicit differentiation. b. Solve the given equation for \(y\) to identify the implicitly defined functions \(y=f_{1}(x), y=f_{2}(x), \ldots\) c. Use the functions found in part (b) to graph the given equation. $$x^{4}=2\left(x^{2}-y^{2}\right) \text { (eight curve) }$$

Let \(C(x)\) represent the cost of producing \(x\) items and \(p(x)\) be the sale price per item if \(x\) items are sold. The profit \(P(x)\) of selling x items is \(P(x)=x p(x)-C(x)\) (revenue minus costs). The average profit per item when \(x\) items are sold is \(P(x) / x\) and the marginal profit is dP/dx. The marginal profit approximates the profit obtained by selling one more item given that \(x\) items have already been sold. Consider the following cost functions \(C\) and price functions \(p\). a. Find the profit function \(P\). b. Find the average profit function and marginal profit function. c. Find the average profit and marginal profit if \(x=a\) units are sold. d. Interpret the meaning of the values obtained in part \((c)\). $$C(x)=-0.02 x^{2}+50 x+100, p(x)=100, a=500$$

Work carefully Proceed with caution when using implicit differentiation to find points at which a curve has a specified slope. For the following curves, find the points on the curve (if they exist) at which the tangent line is horizontal or vertical. Once you have found possible points, make sure they actually lie on the curve. Confirm your results with a graph. $$x\left(1-y^{2}\right)+y^{3}=0$$

Let \(f\) and \(g\) be differentiable functions with \(h(x)=f(g(x)) .\) For a given constant \(a,\) let \(u=g(a)\) and \(v=g(x),\) and define $$H(v)=\left\\{\begin{array}{ll} \frac{f(v)-f(u)}{v-u}-f^{\prime}(u) & \text { if } v \neq u \\ 0 & \text { if } v=u. \end{array}\right.$$ a. Show that \(\lim _{x \rightarrow u} H(v)=0\) b. For any value of \(u\) show that $$f(v)-f(u)=\left(H(v)+f^{\prime}(u)\right)(v-u).$$ c. Show that. $$h^{\prime}(a)=\lim _{x \rightarrow a}\left(\left(H(g(x))+f^{\prime}(g(a))\right) \cdot \frac{g(x)-g(a)}{x-a}\right).$$ d. Show that \(h^{\prime}(a)=f^{\prime}(g(a)) g^{\prime}(a)\).

a. Differentiate both sides of the identity \(\cos 2 t=\cos ^{2} t-\sin ^{2} t\) to prove that \(\sin 2 t=2 \sin t \cos t\). b. Verify that you obtain the same identity for sin \(2 t\) as in part (a) if you differentiate the identity \(\cos 2 t=2 \cos ^{2} t-1\). c. Differentiate both sides of the identity \(\sin 2 t=2 \sin t \cos t\) to prove that \(\cos 2 t=\cos ^{2} t-\sin ^{2} t\).

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