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Sagot :
Sure, let's find the derivative of the function [tex]\( f(x) = 2 \tanh^{-1} \left(\tan \frac{x}{5} \right) \)[/tex].
The function provided is:
[tex]\[ f(x) = 2 \tanh^{-1} \left(\tan \frac{x}{5} \right) \][/tex]
We will first consider the inner function and then apply the chain rule.
1. Inner Function: [tex]\( g(x) = \tanh^{-1} \left(\tan \frac{x}{5} \right) \)[/tex]
As we know, the derivative of [tex]\( \tanh^{-1}(u) \)[/tex] with respect to [tex]\( u \)[/tex] is:
[tex]\[ \frac{d}{du} \tanh^{-1}(u) = \frac{1}{1-u^2} \][/tex]
Here, [tex]\( u = \tan \frac{x}{5} \)[/tex], so:
[tex]\[ \frac{d}{dx} \tanh^{-1} \left(\tan \frac{x}{5} \right) = \frac{1}{1 - \left(\tan \frac{x}{5}\right)^2} \cdot \frac{d}{dx} \left(\tan \frac{x}{5} \right) \][/tex]
2. Derivative of [tex]\( \tan \frac{x}{5} \)[/tex]:
Since the derivative of [tex]\( \tan u \)[/tex] with respect to [tex]\( u \)[/tex] is [tex]\( \sec^2 u \)[/tex], and [tex]\( u = \frac{x}{5} \)[/tex], we have:
[tex]\[ \frac{d}{dx} \left( \tan \frac{x}{5} \right) = \sec^2 \left(\frac{x}{5}\right) \cdot \frac{1}{5} \][/tex]
3. Combining the results:
Putting these together:
[tex]\[ \frac{d}{dx} \tanh^{-1} \left(\tan \frac{x}{5} \right) = \frac{1}{1 - \left(\tan \frac{x}{5}\right)^2} \cdot \frac{1}{5} \sec^2 \left(\frac{x}{5}\right) \][/tex]
4. Multiplication by the outer function:
Finally, since the outer function is [tex]\( 2 \cdot (\text{inner function}) \)[/tex], the derivative will be multiplied by 2:
[tex]\[ f'(x) = 2 \cdot \frac{1}{1 - \left(\tan \frac{x}{5}\right)^2} \cdot \frac{1}{5} \sec^2 \left(\frac{x}{5}\right) \][/tex]
5. Simplifying:
We simplify the result:
[tex]\[ f'(x) = \frac{2 \sec^2 \left(\frac{x}{5}\right)}{5 \left(1 - \left(\tan \frac{x}{5}\right)^2\right)} \][/tex]
Recognizing that [tex]\(\sec^2 u = 1 + \tan^2 u\)[/tex], we can rewrite the expression inside. Hence, considering all combined parts, the final derivative is:
[tex]\[ f'(x) = \frac{2 \left( \tan^2 \left(\frac{x}{5}\right) + 1 \right)}{5 \left( 1 - \left(\tan \frac{x}{5}\right)^2 \right)} \][/tex]
Thus, the derivative of [tex]\( f(x) = 2 \tanh^{-1} \left( \tan \frac{x}{5} \right) \)[/tex] is:
[tex]\[ \boxed{ \frac{2 \left( \tan^2 \left(\frac{x}{5}\right) + 1 \right)}{5 \left( 1 - \left(\tan \frac{x}{5}\right)^2 \right)} } \][/tex]
The function provided is:
[tex]\[ f(x) = 2 \tanh^{-1} \left(\tan \frac{x}{5} \right) \][/tex]
We will first consider the inner function and then apply the chain rule.
1. Inner Function: [tex]\( g(x) = \tanh^{-1} \left(\tan \frac{x}{5} \right) \)[/tex]
As we know, the derivative of [tex]\( \tanh^{-1}(u) \)[/tex] with respect to [tex]\( u \)[/tex] is:
[tex]\[ \frac{d}{du} \tanh^{-1}(u) = \frac{1}{1-u^2} \][/tex]
Here, [tex]\( u = \tan \frac{x}{5} \)[/tex], so:
[tex]\[ \frac{d}{dx} \tanh^{-1} \left(\tan \frac{x}{5} \right) = \frac{1}{1 - \left(\tan \frac{x}{5}\right)^2} \cdot \frac{d}{dx} \left(\tan \frac{x}{5} \right) \][/tex]
2. Derivative of [tex]\( \tan \frac{x}{5} \)[/tex]:
Since the derivative of [tex]\( \tan u \)[/tex] with respect to [tex]\( u \)[/tex] is [tex]\( \sec^2 u \)[/tex], and [tex]\( u = \frac{x}{5} \)[/tex], we have:
[tex]\[ \frac{d}{dx} \left( \tan \frac{x}{5} \right) = \sec^2 \left(\frac{x}{5}\right) \cdot \frac{1}{5} \][/tex]
3. Combining the results:
Putting these together:
[tex]\[ \frac{d}{dx} \tanh^{-1} \left(\tan \frac{x}{5} \right) = \frac{1}{1 - \left(\tan \frac{x}{5}\right)^2} \cdot \frac{1}{5} \sec^2 \left(\frac{x}{5}\right) \][/tex]
4. Multiplication by the outer function:
Finally, since the outer function is [tex]\( 2 \cdot (\text{inner function}) \)[/tex], the derivative will be multiplied by 2:
[tex]\[ f'(x) = 2 \cdot \frac{1}{1 - \left(\tan \frac{x}{5}\right)^2} \cdot \frac{1}{5} \sec^2 \left(\frac{x}{5}\right) \][/tex]
5. Simplifying:
We simplify the result:
[tex]\[ f'(x) = \frac{2 \sec^2 \left(\frac{x}{5}\right)}{5 \left(1 - \left(\tan \frac{x}{5}\right)^2\right)} \][/tex]
Recognizing that [tex]\(\sec^2 u = 1 + \tan^2 u\)[/tex], we can rewrite the expression inside. Hence, considering all combined parts, the final derivative is:
[tex]\[ f'(x) = \frac{2 \left( \tan^2 \left(\frac{x}{5}\right) + 1 \right)}{5 \left( 1 - \left(\tan \frac{x}{5}\right)^2 \right)} \][/tex]
Thus, the derivative of [tex]\( f(x) = 2 \tanh^{-1} \left( \tan \frac{x}{5} \right) \)[/tex] is:
[tex]\[ \boxed{ \frac{2 \left( \tan^2 \left(\frac{x}{5}\right) + 1 \right)}{5 \left( 1 - \left(\tan \frac{x}{5}\right)^2 \right)} } \][/tex]
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