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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]
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