Chapter 16: Problem 3

Find \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) when \(x=1\) where \(y\) is defined by (a) \(\frac{x \cos x}{\sin x}\) (b) \(\frac{x^{2}+1}{x^{x}}\) (c) \(\frac{\sin x \cos 2 x}{x^{2}}\) (d) \(\frac{1}{x \sin x}\) (e) \(\frac{4 \tan x}{\mathrm{e}^{x} \sin x}\)

Short Answer

Expert verified

Given the step-by-step solution, the short answer is: (a) The derivative of \(y=\frac{x\cos x}{\sin x}\) is \(\frac{dy}{dx} = \frac{\sin{x}(\cos{x} - x\sin{x}) - x\cos{x}(\cos{x})}{\sin^2{x}}\) and its value at x=1 is \(\frac{\sin(1)(\cos(1) - 1\sin(1)) - 1\cos(1)(\cos(1))}{\sin^2(1)}\). (b) The derivative of \(y=\frac{x^2+1}{x^x}\) is \(\frac{dy}{dx} = \frac{2x(x^x) - (x^2+1)(x^x(\ln{x} + 1))}{x^{2x}}\) and its value at x=1 is 0. (c) The derivative of \(y=\frac{\sin x \cos 2 x}{x^{2}}\) is \(\frac{dy}{dx} = \frac{2x(\cos{x}\cos{2x} - 2\sin^2{x}) - (\sin{x}\cos{2x})(2x)}{x^4}\) and its value at x=1 is \(\frac{2(1)(\cos(1)\cos(2) - 2\sin^2(1)) - (\sin(1)\cos(2))(2(1))}{1^4}\). (d) The derivative of \(y=\frac{1}{x\sin x}\) is \(\frac{dy}{dx} = \frac{-(\cos{x} + x\cos{x})}{x^2\sin^2{x}}\) and its value at x=1 is \(\frac{-(\cos(1) + 1\cos(1))}{1^2\sin^2(1)}\). (e) The derivative of \(y=\frac{4\tan x}{e^x\sin x}\) is \(\frac{dy}{dx} = \frac{(e^x\sin{x})(4\sec^2{x}) - (4\tan{x})((\cos{x})e^{x} + \sin{x}e^{x})}{(e^x\sin{x})^2}\) and its value at x=1 is \(\frac{(e\sin{1})(4\sec^2{1}) - (4\tan{1})((\cos{1})e + \sin{1}e)}{(e\sin{1})^2}\).

Step by step solution

(a) Differentiate \(y=\frac{x \cos x}{\sin x}\)

We'll use the quotient rule, which says \(\frac{d}{dx}(\frac{u}{v}) = \frac{vu' - uv'}{v^2}\). Let \(u = x\cos{x}\) and \(v = \sin{x}\): \(\frac{d}{dx}(u) = \frac{d}{dx}(x\cos{x}) = \cos{x} - x\sin{x}\) using product rule. \(\frac{d}{dx}(v) = \frac{d}{dx}(\sin{x}) = \cos{x}\). Now, apply the quotient rule: \(\frac{dy}{dx} = \frac{\sin{x}(\cos{x} - x\sin{x}) - x\cos{x}(\cos{x})}{\sin^2{x}}\)

Evaluate the derivative at x=1

To find the derivative at x=1, substitute x=1 in the above expression: \(\frac{dy}{dx} \vert_{x=1} = \frac{\sin(1)(\cos(1) - 1\sin(1)) - 1\cos(1)(\cos(1))}{\sin^2(1)}\)

(b) Differentiate \(y=\frac{x^2+1}{x^x}\)

Using the quotient rule with \(u = x^2+1\) and \(v = x^x\): \(\frac{d}{dx}(u) = \frac{d}{dx}(x^2+1) = 2x\) \(\frac{d}{dx}(v) = \frac{d}{dx}(x^x) = x^x(\ln{x} + 1)\) using the derivative of \(a^x\) Now, apply the quotient rule: \(\frac{dy}{dx} = \frac{2x(x^x) - (x^2+1)(x^x(\ln{x} + 1))}{x^{2x}}\)

Evaluate the derivative at x=1

To find the derivative at x=1, substitute x=1 in the above expression: \(\frac{dy}{dx} \vert_{x=1} = \frac{2(1^{1}) - (1^2+1)(1^1(\ln{1} + 1))}{1^{2}} = 0\)

(c) Differentiate \(y=\frac{\sin x \cos 2 x}{x^{2}}\)

Using the quotient rule with \(u = \sin{x}\cos{2x}\) and \(v = x^2\): \(\frac{d}{dx}(u) = \frac{d}{dx}(\sin{x}\cos{2x}) = \cos{x}\cos{2x} - 2\sin^2{x}\) using product rule and chain rule. \(\frac{d}{dx}(v) = \frac{d}{dx}(x^2) = 2x\). Now, apply the quotient rule: \(\frac{dy}{dx} = \frac{2x(\cos{x}\cos{2x} - 2\sin^2{x}) - (\sin{x}\cos{2x})(2x)}{x^4}\)

Evaluate the derivative at x=1

To find the derivative at x=1, substitute x=1 in the above expression: \(\frac{dy}{dx} \vert_{x=1} = \frac{2(1)(\cos(1)\cos(2) - 2\sin^2(1)) - (\sin(1)\cos(2))(2(1))}{1^4}\)

(d) Differentiate \(y=\frac{1}{x \sin x}\)

Using the quotient rule with \(u = 1\) and \(v = x\sin{x}\): \(\frac{d}{dx}(u) = \frac{d}{dx}(1) = 0\) \(\frac{d}{dx}(v) = \frac{d}{dx}(x\sin{x}) = \cos{x} + x\cos{x}\) using product rule. Now, apply the quotient rule: \(\frac{dy}{dx} = \frac{-(\cos{x} + x\cos{x})}{x^2\sin^2{x}}\)

Evaluate the derivative at x=1

To find the derivative at x=1, substitute x=1 in the above expression: \(\frac{dy}{dx} \vert_{x=1} = \frac{-(\cos(1) + 1\cos(1))}{1^2\sin^2(1)}\)

(e) Differentiate \(y=\frac{4 \tan x}{e^{x} \sin x}\)

Using the quotient rule with \(u = 4\tan{x}\) and \(v = e^{x}\sin{x}\): \(\frac{d}{dx}(u) = \frac{d}{dx}(4\tan{x}) = 4\sec^2{x}\) \(\frac{d}{dx}(v) = \frac{d}{dx}(e^{x}\sin{x}) = (\cos{x})e^{x} + \sin{x}e^{x}\) using product rule. Now, apply the quotient rule: \(\frac{dy}{dx} = \frac{(e^x\sin{x})(4\sec^2{x}) - (4\tan{x})((\cos{x})e^{x} + \sin{x}e^{x})}{(e^x\sin{x})^2}\)

Evaluate the derivative at x=1

To find the derivative at x=1, substitute x=1 in the above expression: \(\frac{dy}{dx} \vert_{x=1} = \frac{(e\sin{1})(4\sec^2{1}) - (4\tan{1})((\cos{1})e + \sin{1}e)}{(e\sin{1})^2}\)

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Key Concepts

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

The Quotient Rule

In calculus, the quotient rule is a method for finding the derivative of a function that is the ratio of two differentiable functions. When faced with a function that is the division of one expression by another, the quotient rule provides a systematic way to differentiate it. The rule states that for two functions, where one is divided by the other, \(u/v\), the derivative, \(u'/v'\), is given by \(\frac{vu' - uv'}{v^2}\).

Consider the function \(y = \frac{x \cos x}{\sin x}\). Applying the quotient rule, we first differentiate the numerator \(u = x\cos{x}\) to get \(\cos{x} - x\sin{x}\), using the product rule, and the denominator \(v = \sin{x}\) to get \(\cos{x}\). We then apply the quotient rule to find the derivative of the entire expression. It is crucial to simplify the result as much as possible for clearer understanding and further evaluation at specific points.

The Product Rule

The product rule is another fundamental technique in calculus used to differentiate functions that are being multiplied together. Given two differentiable functions \(u(x)\) and \(v(x)\), the derivative of their product \(u(x)\cdot v(x)\) is given by \(u'(x)v(x) + u(x)v'(x)\).

This rule is particularly useful when dealing with trigonometric functions multiplied by algebraic expressions, as in the exercise where we differentiate the numerator \(u = \sin{x}\cos{2x}\). We use the product rule here because we are dealing with the derivative of a product of two functions of \(x\). The key is to differentiate each function separately and then apply the product rule to combine the derivatives in the correct formula.

The Chain Rule

The chain rule is a powerful tool used to differentiate composite functions, where one function is nested inside another. This rule allows us to find the derivative of the outer function with respect to the inner function, and then multiply that by the derivative of the inner function with respect to its variable. Formally, for \(y = f(g(x))\), the derivative \(dy/dx\) is \(f'(g(x))\cdot g'(x)\).

When we encounter a function like \(y = \sin{x}\cos{2x}\), and we need to apply the derivative of \(\cos{2x}\), we utilize the chain rule because \(2x\) is a function inside another function, \(\cos{x}\). Thorough comprehension and application of the chain rule are essential when handling more complex differentiation problems.

Derivatives of Exponential Functions

Derivatives of exponential functions introduce the concept of differentiating functions with a base of \(e\), the natural logarithm base. The derivative of an exponential function \(e^{ax}\) with respect to \(x\) is \(ae^{ax}\).

This concept is pivotal when you consider functions such as \(y = \frac{4 \tan x}{e^{x} \sin x}\). By understanding the differentiation of exponential functions, we can apply the product and quotient rules more effectively, since terms involving \(e^{x}\) will frequently appear in the numerators or denominators of our functions.

Evaluating Derivatives at a Point

Once the derivative of a function is known, the next step might involve finding the slope of the tangent line to the function's graph at a certain point. This process is known as evaluating the derivative at a point. It simply involves substituting the point's \(x\)-value into the derivative function to get the slope at that point.

For example, after finding the derivative of \(y = \frac{x \cos x}{\sin x}\), evaluating it at \(x = 1\) gives us the rate of change of the function at that specific point. This step is essential in many applications, such as finding the instantaneous rate of change of a physical process or determining the speed of an object at a specific time.

Find \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) when \(x=1\) where \(y\) is defined by (a) \(\frac{x \cos x}{\sin x}\) (b) \(\frac{x^{2}+1}{x^{x}}\) (c) \(\frac{\sin x \cos 2 x}{x^{2}}\) (d) \(\frac{1}{x \sin x}\) (e) \(\frac{4 \tan x}{\mathrm{e}^{x} \sin x}\)

Short Answer

Step by step solution

(a) Differentiate \(y=\frac{x \cos x}{\sin x}\)

Evaluate the derivative at x=1

(b) Differentiate \(y=\frac{x^2+1}{x^x}\)

Evaluate the derivative at x=1

(c) Differentiate \(y=\frac{\sin x \cos 2 x}{x^{2}}\)

Evaluate the derivative at x=1

(d) Differentiate \(y=\frac{1}{x \sin x}\)

Evaluate the derivative at x=1

(e) Differentiate \(y=\frac{4 \tan x}{e^{x} \sin x}\)

Evaluate the derivative at x=1

Key Concepts

The Quotient Rule

The Product Rule

The Chain Rule

Derivatives of Exponential Functions

Evaluating Derivatives at a Point

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