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Chapter 3: Problem 36

Find the derivative of the following functions. $$y=\frac{\tan w}{1+\tan w}$$

Short Answer

Expert verified
Answer: The derivative of the function is \(y' = \frac{\sec^2 w}{(1 + \tan w)^2}\).

Step by step solution

01

Identify the necessary rules

In this case, we need to use the quotient rule and the chain rule for differentiation. The quotient rule states that the derivative of \(\frac{u}{v}\) is \(\frac{vu' - uv'}{v^2}\). The Chain rule states that the derivative of f(g(x)) is f'(g(x)) * g'(x).
02

Apply the quotient rule

Applying the quotient rule, we'll find the derivative \(y'\) of the function \(y = \frac{\tan w}{1 + \tan w}\), where \(u = \tan w\) and \(v = 1 + \tan w\). Therefore, we have: $$y' = \frac{v \cdot u' - u \cdot v'}{v^2}$$
03

Find the derivatives of u and v

We will now find the derivatives \(u'(w)\) and \(v'(w)\). To do this, we need to use the chain rule: 1. For \(u(w) = \tan w\), \(u'(w) = \frac{d}{dw}(\tan w) = \sec^2 w\) 2. For \(v(w) = 1 + \tan w\), \(v'(w) = \frac{d}{dw}(1 + \tan w) = 0 + \frac{d}{dw}(\tan w) = \sec^2 w\)
04

Substitute the derivatives of u and v into the quotient rule formula

Now, we'll substitute \(u'(w) = \sec^2 w\) and \(v'(w) = \sec^2 w\) into the formula from Step 2: $$y' = \frac{(1 + \tan w) \cdot \sec^2 w - (\tan w) \cdot \sec^2 w}{(1 + \tan w)^2}$$
05

Simplify the expression

Now, we will simplify the expression derived in Step 4: $$y' = \frac{\sec^2 w(1 + \tan w - \tan w)}{(1 + \tan w)^2} = \frac{\sec^2 w}{(1 + \tan w)^2}$$ So, the derivative of the function \(y = \frac{\tan w}{1 + \tan w}\) is: $$y' = \frac{\sec^2 w}{(1 + \tan w)^2}$$

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Most popular questions from this chapter

Suppose a large company makes 25,000 gadgets per year in batches of \(x\) items at a time. After analyzing setup costs to produce each batch and taking into account storage costs, it has been determined that the total cost \(C(x)\) of producing 25,000 gadgets in batches of \(x\) items at a time is given by $$C(x)=1,250,000+\frac{125,000,000}{x}+1.5 x.$$ a. Determine the marginal cost and average cost functions. Graph and interpret these functions. b. Determine the average cost and marginal cost when \(x=5000\). c. The meaning of average cost and marginal cost here is different from earlier examples and exercises. Interpret the meaning of your answer in part (b).

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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 \((c)\). $$C(x)=-0.02 x^{2}+50 x+100, p(x)=100, a=500$$

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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