Differentiate coth(tan(x)) with respect to x.
Understand the Problem
The question is asking us to find the derivative of the hyperbolic cotangent of the tangent function with respect to x. To solve this, we will apply the chain rule and differentiation rules to find the derivative of coth(tan(x)).
Answer
The derivative is given by: $$ \frac{dy}{dx} = -\text{csch}^2(\tan(x)) \cdot \sec^2(x) $$
Answer for screen readers
The derivative of $\text{coth}(\tan(x))$ is given by:
$$ \frac{dy}{dx} = -\text{csch}^2(\tan(x)) \cdot \sec^2(x) $$
Steps to Solve
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Identify the function
The function we need to differentiate is $y = \text{coth}(\tan(x))$. -
Apply the chain rule
Using the chain rule, we differentiate $y$ with respect to $x$. The chain rule states that if you have a composite function $f(g(x))$, its derivative is given by: $$ \frac{dy}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx} $$ -
Differentiate the outer function
First, we differentiate the outer function, which is $\text{coth}(u)$, where $u = \tan(x)$. The derivative of $\text{coth}(u)$ is: $$ \frac{d}{du} \text{coth}(u) = -\text{csch}^2(u) $$ -
Differentiate the inner function
Next, we differentiate the inner function $u = \tan(x)$. The derivative of $\tan(x)$ is: $$ \frac{du}{dx} = \sec^2(x) $$ -
Combine the derivatives
Now we combine these derivatives using the chain rule: $$ \frac{dy}{dx} = -\text{csch}^2(\tan(x)) \cdot \sec^2(x) $$ -
Final result
Thus, the derivative of the function is: $$ \frac{dy}{dx} = -\text{csch}^2(\tan(x)) \cdot \sec^2(x) $$
The derivative of $\text{coth}(\tan(x))$ is given by:
$$ \frac{dy}{dx} = -\text{csch}^2(\tan(x)) \cdot \sec^2(x) $$
More Information
The hyperbolic cotangent function, $\text{coth}(x)$, is related to the hyperbolic sine and cosine functions. The derivative of the tangent function is crucial in this problem, highlighting the importance of both trigonometric and hyperbolic functions in calculus.
Tips
- Forgetting to apply the chain rule properly can lead to incorrect results. Ensure that both the inner and outer functions are differentiated correctly.
- Confusing the derivatives of hyperbolic functions with trigonometric functions can also lead to mistakes.
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