Question

Compute the derivative of the given function. $$h(x)=\cot x-e^{x}$$

Short answer

The derivative is \(-\csc^2 x - e^x\).

Step-by-step solution

Identify the Derivatives

To find the derivative of the given function, we identify the derivatives of each individual term in the function. The function is given by: \( h(x) = \cot x - e^x \). The derivative of \( \cot x \) is \(-\csc^2 x \) and the derivative of \( e^x \) is \( e^x \).

Apply the Sum/Difference Rule

The derivative of a sum or difference of functions is the sum or difference of their derivatives. Apply this rule to the function \( h(x)=\cot x-e^x \) by differentiating each term separately and then combining the results.

Compute the Derivative

Differentiate each term separately and combine: \( h'(x) = \frac{d}{dx}(\cot x) - \frac{d}{dx}(e^x) = -\csc^2 x - e^x \).

Write the Final Answer

So, the derivative of the function \( h(x)=\cot x-e^x \) is \( h'(x) = -\csc^2 x - e^x \).

Key concepts

Trigonometric Derivatives

Understanding trigonometric derivatives is crucial for calculus. Trigonometric functions, like \( \cot x \), have specific derivatives that help us analyze changes in angles and cycles. The cotangent function, \( \cot x = \frac{\cos x}{\sin x} \), is an example. Its derivative, \( \frac{d}{dx}(\cot x) = -\csc^2 x \), is derived using the quotient rule and knowledge of the derivatives of sine and cosine functions.

• For \( \cot x \), you use its identity, \( \csc x = \frac{1}{\sin x} \), to obtain \(-\csc^2 x\).
• Recognize that trigonometric derivatives come up often, so memorizing them can be helpful.

When differentiating, always observe the given function's components and apply the right derivative formulas, especially for trigonometric expressions.

Sum and Difference Rule

The sum and difference rule in calculus simplifies the differentiation process when dealing with sums or differences of functions. Essentially, it states that the derivative of a sum is the sum of the derivatives, and similarly for differences. This rule allows us to handle complex expressions easily.

• Suppose you have a function like \( h(x) = \cot x - e^x \).
• You differentiate \( \cot x \) to get \( -\csc^2 x \), and \( e^x \) to get \( e^x \).
• Combine these individual derivatives by subtracting: \( h'(x) = -\csc^2 x - e^x \).

This rule is a fundamental tool that saves time and reduces errors when calculating derivatives in more complex scenarios.

Exponential Functions

Exponential functions, like \( e^x \), are a key part of mathematical analysis because of their growth properties and appearances in natural phenomena. The derivative of the exponential function \( e^x \) is quite unique and straightforward, as it is the only function whose rate of change is equal to itself: \( \frac{d}{dx}(e^x) = e^x \).
This property makes exponential functions straightforward to work with in calculus problems, especially when combined with other types of functions.

• In mixed function problems, such as with trigonometric terms, it's important to apply their derivatives correctly.
• Always remember the simple fact that the exponential function's derivative doesn't change its form, which can simplify many calculations.

Understanding how to differentiate exponential functions paves the way to handling more complex expression derivatives efficiently.