Get the answers you need at Westonci.ca, where our expert community is dedicated to providing you with accurate information. Discover precise answers to your questions from a wide range of experts on our user-friendly Q&A platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
Sure, let's find the first and second derivatives of the function [tex]\( h(x) = \tan(3x + 5) \)[/tex] step by step.
### Step 1: First Derivative [tex]\( h'(x) \)[/tex]
First, recall the derivative of the tangent function:
[tex]\[ \frac{d}{dx} [\tan(u)] = \sec^2(u) \cdot \frac{du}{dx} \][/tex]
Given [tex]\( h(x) = \tan(3x + 5) \)[/tex], let's set [tex]\( u = 3x + 5 \)[/tex]. Then we need to find the derivative of [tex]\( u \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{du}{dx} = \frac{d}{dx} [3x + 5] = 3 \][/tex]
Using the chain rule:
[tex]\[ h'(x) = \frac{d}{dx} [\tan(3x + 5)] = \sec^2(3x + 5) \cdot \frac{d}{dx} [3x + 5] = \sec^2(3x + 5) \cdot 3 \][/tex]
Recall the trigonometric identity:
[tex]\[ \sec^2(u) = 1 + \tan^2(u) \][/tex]
Thus, we can express the first derivative as:
[tex]\[ h'(x) = 3 \sec^2(3x + 5) = 3 (1 + \tan^2(3x + 5)) \][/tex]
Since [tex]\( h(x) = \tan(3x + 5) \)[/tex], we have:
[tex]\[ h'(x) = 3 (1 + \tan^2(3x + 5)) \][/tex]
This simplifies to:
[tex]\[ h'(x) = 3 (\tan^2(3x + 5) + 1) \][/tex]
Finally:
[tex]\[ h'(x) = 3\tan^2(3x + 5) + 3 \][/tex]
### Step 2: Second Derivative [tex]\( h''(x) \)[/tex]
Now, let's move on to the second derivative [tex]\( h''(x) \)[/tex]. Using the result from [tex]\( h'(x) \)[/tex], we get:
[tex]\[ h'(x) = 3\tan^2(3x + 5) + 3 \][/tex]
We need to differentiate [tex]\( h'(x) \)[/tex] again with respect to [tex]\( x \)[/tex]. Let's break it into two parts.
1. Differentiate [tex]\( 3\tan^2(3x + 5) \)[/tex]:
[tex]\[ \frac{d}{dx} [3\tan^2(3x + 5)] = 3 \cdot 2\tan(3x + 5) \cdot \frac{d}{dx} [\tan(3x + 5)] \][/tex]
Now, we need [tex]\( \frac{d}{dx} [\tan(3x + 5)] \)[/tex]:
[tex]\[ \frac{d}{dx} [\tan(3x + 5)] = \sec^2(3x + 5) \cdot 3 \][/tex]
Therefore:
[tex]\[ 6\tan(3x + 5) \cdot 3\sec^2(3x + 5) = 18\tan(3x + 5) \sec^2(3x + 5) \][/tex]
2. Differentiate the constant term [tex]\( 3 \)[/tex], which is [tex]\( 0 \)[/tex].
Combining these results, the second derivative is:
[tex]\[ h''(x) = 18\tan(3x + 5) \sec^2(3x + 5) \][/tex]
Recall:
[tex]\[ \sec^2(u) = 1 + \tan^2(u) \][/tex]
[tex]\[ \sec^2(3x + 5) = 1 + \tan^2(3x + 5) \][/tex]
Now substitute back:
[tex]\[ h''(x) = 18\tan(3x + 5) (1 + \tan^2(3x + 5)) \][/tex]
Simplifying:
[tex]\[ h''(x) = 18\tan(3x + 5)(1 + \tan^2(3x + 5)) = 18\tan(3x + 5) (1 + \tan^2(3x + 5)) \][/tex]
Since [tex]\( 6 \times 3 = 18 \)[/tex], we can write it as:
[tex]\[ h''(x) = 3 (6 \tan(3x + 5)^2 + 6) \tan(3x + 5) \][/tex]
However, preserving the simplified form:
[tex]\[ h''(x) = 3 (6\tan^2(3x + 5) + 6) \tan(3x + 5) \][/tex]
Hence, the final answers are:
[tex]\[ \begin{array}{l} h^{\prime}(x) = 3\tan^2(3x + 5) + 3 \\ h^{\prime \prime}(x) = 3 (6\tan^2(3x + 5) + 6) \tan(3x + 5) \end{array} \][/tex]
### Step 1: First Derivative [tex]\( h'(x) \)[/tex]
First, recall the derivative of the tangent function:
[tex]\[ \frac{d}{dx} [\tan(u)] = \sec^2(u) \cdot \frac{du}{dx} \][/tex]
Given [tex]\( h(x) = \tan(3x + 5) \)[/tex], let's set [tex]\( u = 3x + 5 \)[/tex]. Then we need to find the derivative of [tex]\( u \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{du}{dx} = \frac{d}{dx} [3x + 5] = 3 \][/tex]
Using the chain rule:
[tex]\[ h'(x) = \frac{d}{dx} [\tan(3x + 5)] = \sec^2(3x + 5) \cdot \frac{d}{dx} [3x + 5] = \sec^2(3x + 5) \cdot 3 \][/tex]
Recall the trigonometric identity:
[tex]\[ \sec^2(u) = 1 + \tan^2(u) \][/tex]
Thus, we can express the first derivative as:
[tex]\[ h'(x) = 3 \sec^2(3x + 5) = 3 (1 + \tan^2(3x + 5)) \][/tex]
Since [tex]\( h(x) = \tan(3x + 5) \)[/tex], we have:
[tex]\[ h'(x) = 3 (1 + \tan^2(3x + 5)) \][/tex]
This simplifies to:
[tex]\[ h'(x) = 3 (\tan^2(3x + 5) + 1) \][/tex]
Finally:
[tex]\[ h'(x) = 3\tan^2(3x + 5) + 3 \][/tex]
### Step 2: Second Derivative [tex]\( h''(x) \)[/tex]
Now, let's move on to the second derivative [tex]\( h''(x) \)[/tex]. Using the result from [tex]\( h'(x) \)[/tex], we get:
[tex]\[ h'(x) = 3\tan^2(3x + 5) + 3 \][/tex]
We need to differentiate [tex]\( h'(x) \)[/tex] again with respect to [tex]\( x \)[/tex]. Let's break it into two parts.
1. Differentiate [tex]\( 3\tan^2(3x + 5) \)[/tex]:
[tex]\[ \frac{d}{dx} [3\tan^2(3x + 5)] = 3 \cdot 2\tan(3x + 5) \cdot \frac{d}{dx} [\tan(3x + 5)] \][/tex]
Now, we need [tex]\( \frac{d}{dx} [\tan(3x + 5)] \)[/tex]:
[tex]\[ \frac{d}{dx} [\tan(3x + 5)] = \sec^2(3x + 5) \cdot 3 \][/tex]
Therefore:
[tex]\[ 6\tan(3x + 5) \cdot 3\sec^2(3x + 5) = 18\tan(3x + 5) \sec^2(3x + 5) \][/tex]
2. Differentiate the constant term [tex]\( 3 \)[/tex], which is [tex]\( 0 \)[/tex].
Combining these results, the second derivative is:
[tex]\[ h''(x) = 18\tan(3x + 5) \sec^2(3x + 5) \][/tex]
Recall:
[tex]\[ \sec^2(u) = 1 + \tan^2(u) \][/tex]
[tex]\[ \sec^2(3x + 5) = 1 + \tan^2(3x + 5) \][/tex]
Now substitute back:
[tex]\[ h''(x) = 18\tan(3x + 5) (1 + \tan^2(3x + 5)) \][/tex]
Simplifying:
[tex]\[ h''(x) = 18\tan(3x + 5)(1 + \tan^2(3x + 5)) = 18\tan(3x + 5) (1 + \tan^2(3x + 5)) \][/tex]
Since [tex]\( 6 \times 3 = 18 \)[/tex], we can write it as:
[tex]\[ h''(x) = 3 (6 \tan(3x + 5)^2 + 6) \tan(3x + 5) \][/tex]
However, preserving the simplified form:
[tex]\[ h''(x) = 3 (6\tan^2(3x + 5) + 6) \tan(3x + 5) \][/tex]
Hence, the final answers are:
[tex]\[ \begin{array}{l} h^{\prime}(x) = 3\tan^2(3x + 5) + 3 \\ h^{\prime \prime}(x) = 3 (6\tan^2(3x + 5) + 6) \tan(3x + 5) \end{array} \][/tex]
We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.
