Looking for answers? Westonci.ca is your go-to Q&A platform, offering quick, trustworthy responses from a community of experts. Explore our Q&A platform to find reliable answers from a wide range of experts in different fields. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
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 our answers were helpful. Return anytime for more information and answers to any other questions you may have. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.
