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Chapter 3: Problem 41

Use the Chain Rule to find the derivative of the following functions. $$y=(1+2 \tan x)^{15}$$

Short Answer

Expert verified
Answer: The derivative of the function \(y=(1+2 \tan x)^{15}\) with respect to \(x\) is \(\frac{dy}{dx} = 30(1+2\tan x)^{14}\sec^2 x\).

Step by step solution

01

Identify the outer and inner functions

Let's identify the outer function and the inner function of our given function: Outer function: \(f(u) = u^{15}\) where \(u\) is the inner function (\(u = 1 +2\tan x\)) Inner function: \(g(x) = 1 + 2 \tan x\)
02

Find the derivatives of the outer and inner functions

Find the derivative of the outer function with respect to its inner function. Using the power rule, we get: $$f'(u) = 15u^{14}$$ Now, find the derivative of the inner function with respect to \(x\). Using the sum rule and the derivative of the tangent function (\(\frac{d}{dx}(\tan x) = \sec^2 x\)), we get: $$g'(x) = 0 + 2\sec^2{x} = 2 \sec^2 x$$
03

Apply the Chain Rule

Now, we can apply the Chain Rule to find the derivative of the given function. The Chain Rule states: $$\frac{dy}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx}$$ Substitute the expressions we obtained in Step 2: $$\frac{dy}{dx} = (15u^{14}) \cdot (2 \sec^2 x)$$ Finally, replace \(u\) with the inner function (\(1 + 2\tan x\)): $$\frac{dy}{dx} = 15(1+2\tan x)^{14} \cdot (2 \sec^2 x)$$
04

Simplify the expression

Our final expression for the derivative is: $$\frac{dy}{dx} = 30(1+2\tan x)^{14}\sec^2 x$$ So, the derivative of the function \(y=(1+2 \tan x)^{15}\) with respect to \(x\) is: $$\frac{dy}{dx} = 30(1+2\tan x)^{14}\sec^2 x$$

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Most popular questions from this chapter

Use a trigonometric identity to show that the derivatives of the inverse cotangent and inverse cosecant differ from the derivatives of the inverse tangent and inverse secant, respectively, by a multiplicative factor of -1

Graphing with inverse trigonometric functions a. Graph the function \(f(x)=\frac{\tan ^{-1} x}{x^{2}+1}\) b. Compute and graph \(f^{\prime}\) and determine (perhaps approximately) the points at which \(f^{\prime}(x)=0\) c. Verify that the zeros of \(f^{\prime}\) correspond to points at which \(f\) has a horizontal tangent line.

Special Product Rule In general, the derivative of a product is not the product of the derivatives. Find nonconstant functions \(f\) and \(g\) such that the derivative of \(f g\) equals \(f^{\prime} g^{\prime}\).

A lighthouse stands 500 m off of a straight shore, the focused beam of its light revolving four times each minute. As shown in the figure, \(P\) is the point on shore closest to the lighthouse and \(Q\) is a point on the shore 200 m from \(P\). What is the speed of the beam along the shore when it strikes the point \(Q ?\) Describe how the speed of the beam along the shore varies with the distance between \(P\) and \(Q\). Neglect the height of the lighthouse.

Find \(f^{\prime}(x), f^{\prime \prime}(x),\) and \(f^{\prime \prime \prime}(x)\) \(f(x)=\frac{x}{x+2}\)
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