Step-by-Step Guidance for Derivative Formulas and Rules

Q1. What is the derivative of $\csc^{-1}(u)$ with respect to $x$?

Background

Topic: Derivatives of Inverse Trigonometric Functions

This question tests your knowledge of how to differentiate the inverse cosecant function, where $u$ is a differentiable function of $x$.

Key Terms and Formulas

• $\csc^{-1}(u)$: Inverse cosecant function

• Chain Rule: $\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$

Step-by-Step Guidance

1. Recall the general formula for the derivative of $\csc^{-1}(u)$: $\frac{d}{du}\csc^{-1}(u)$.

2. Apply the chain rule since $u$ is a function of $x$.

3. Set up the formula using the chain rule, but do not compute the final answer yet.

Try solving on your own before revealing the answer!

Final Answer:

This formula uses the chain rule and the standard derivative for the inverse cosecant function.

Q2. What is the derivative of $\cos(u)$ with respect to $x$?

Background

Topic: Derivatives of Trigonometric Functions

This question tests your ability to differentiate the cosine function, where $u$ is a differentiable function of $x$.

Key Terms and Formulas

• $\cos(u)$: Cosine function

• Chain Rule

Step-by-Step Guidance

1. Recall the basic derivative: $\frac{d}{du}\cos(u) = -\sin(u)$.

2. Apply the chain rule to account for $u$ being a function of $x$.

3. Set up the expression for the derivative, but do not compute the final answer yet.

Try solving on your own before revealing the answer!

Final Answer:

$\frac{d}{dx}\cos(u) = -\sin(u) \cdot \frac{du}{dx}$

The chain rule multiplies the derivative of the outer function by the derivative of the inner function.

Q3. What is the derivative of $\csc(u)$ with respect to $x$?

Background

Topic: Derivatives of Trigonometric Functions

This question tests your ability to differentiate the cosecant function, where $u$ is a differentiable function of $x$.

Key Terms and Formulas

• $\csc(u)$: Cosecant function

• Chain Rule

Step-by-Step Guidance

1. Recall the basic derivative: $\frac{d}{du}\csc(u) = -\csc(u)\cot(u)$.

2. Apply the chain rule for $u$ as a function of $x$.

3. Set up the expression for the derivative, but do not compute the final answer yet.

Try solving on your own before revealing the answer!

Final Answer:

$\frac{d}{dx}\csc(u) = -\csc(u)\cot(u) \cdot \frac{du}{dx}$

The chain rule is applied to the standard derivative of the cosecant function.

Q4. What is the derivative of $\sin(u)$ with respect to $x$?

Background

Topic: Derivatives of Trigonometric Functions

This question tests your ability to differentiate the sine function, where $u$ is a differentiable function of $x$.

Key Terms and Formulas

• $\sin(u)$: Sine function

• Chain Rule

Step-by-Step Guidance

1. Recall the basic derivative: $\frac{d}{du}\sin(u) = \cos(u)$.

2. Apply the chain rule for $u$ as a function of $x$.

3. Set up the expression for the derivative, but do not compute the final answer yet.

Try solving on your own before revealing the answer!

Final Answer:

$\frac{d}{dx}\sin(u) = \cos(u) \cdot \frac{du}{dx}$

The chain rule multiplies the derivative of the outer function by the derivative of the inner function.

Q5. What is the derivative of $\sec(u)$ with respect to $x$?

Background

Topic: Derivatives of Trigonometric Functions

This question tests your ability to differentiate the secant function, where $u$ is a differentiable function of $x$.

Key Terms and Formulas

• $\sec(u)$: Secant function

• Chain Rule

Step-by-Step Guidance

1. Recall the basic derivative: $\frac{d}{du}\sec(u) = \sec(u)\tan(u)$.

2. Apply the chain rule for $u$ as a function of $x$.

3. Set up the expression for the derivative, but do not compute the final answer yet.

Try solving on your own before revealing the answer!

Final Answer:

$\frac{d}{dx}\sec(u) = \sec(u)\tan(u) \cdot \frac{du}{dx}$

The chain rule is applied to the standard derivative of the secant function.

Q6. What is the derivative of $\cot(u)$ with respect to $x$?

Background

Topic: Derivatives of Trigonometric Functions

This question tests your ability to differentiate the cotangent function, where $u$ is a differentiable function of $x$.

Key Terms and Formulas

• $\cot(u)$: Cotangent function

• Chain Rule

Step-by-Step Guidance

1. Recall the basic derivative: $\frac{d}{du}\cot(u) = -\csc^2(u)$.

2. Apply the chain rule for $u$ as a function of $x$.

3. Set up the expression for the derivative, but do not compute the final answer yet.

Try solving on your own before revealing the answer!

Final Answer:

$\frac{d}{dx}\cot(u) = -\csc^2(u) \cdot \frac{du}{dx}$

The chain rule multiplies the derivative of the outer function by the derivative of the inner function.

Q7. What is the derivative of $\log_a(u)$ with respect to $x$?

Background

Topic: Derivatives of Logarithmic Functions

This question tests your ability to differentiate logarithmic functions with a base $a$, where $u$ is a differentiable function of $x$.

Key Terms and Formulas

• $\log_a(u)$: Logarithm base $a$

• Chain Rule

Step-by-Step Guidance

1. Recall the basic derivative: $\frac{d}{du}\log_a(u) = \frac{1}{u \ln(a)}$.

2. Apply the chain rule for $u$ as a function of $x$.

3. Set up the expression for the derivative, but do not compute the final answer yet.

Try solving on your own before revealing the answer!

Final Answer:

$\frac{d}{dx}\log_a(u) = \frac{1}{u \ln(a)} \cdot \frac{du}{dx}$

The chain rule is applied to the standard derivative of the logarithm function.

Q8. What is the derivative of $e^u$ with respect to $x$?

Background

Topic: Derivatives of Exponential Functions

This question tests your ability to differentiate the exponential function, where $u$ is a differentiable function of $x$.

Key Terms and Formulas

• $e^u$: Exponential function

• Chain Rule

Step-by-Step Guidance

1. Recall the basic derivative: $\frac{d}{du}e^u = e^u$.

2. Apply the chain rule for $u$ as a function of $x$.

3. Set up the expression for the derivative, but do not compute the final answer yet.

Try solving on your own before revealing the answer!

Final Answer:

$\frac{d}{dx}e^u = e^u \cdot \frac{du}{dx}$

The chain rule multiplies the derivative of the outer function by the derivative of the inner function.

Q9. What is the derivative of $\tan^{-1}(u)$ with respect to $x$?

Background

Topic: Derivatives of Inverse Trigonometric Functions

This question tests your ability to differentiate the inverse tangent function, where $u$ is a differentiable function of $x$.

Key Terms and Formulas

• $\tan^{-1}(u)$: Inverse tangent function

• Chain Rule

Step-by-Step Guidance

1. Recall the basic derivative: $\frac{d}{du}\tan^{-1}(u) = \frac{1}{1+u^2}$.

2. Apply the chain rule for $u$ as a function of $x$.

3. Set up the expression for the derivative, but do not compute the final answer yet.

Try solving on your own before revealing the answer!

Final Answer:

$\frac{d}{dx}\tan^{-1}(u) = \frac{1}{1+u^2} \cdot \frac{du}{dx}$

The chain rule is applied to the standard derivative of the inverse tangent function.

Q10. What is the derivative of $\cot^{-1}(u)$ with respect to $x$?

Background

Topic: Derivatives of Inverse Trigonometric Functions

This question tests your ability to differentiate the inverse cotangent function, where $u$ is a differentiable function of $x$.

Key Terms and Formulas

• $\cot^{-1}(u)$: Inverse cotangent function

• Chain Rule

Step-by-Step Guidance

1. Recall the basic derivative: $\frac{d}{du}\cot^{-1}(u) = -\frac{1}{1+u^2}$.

2. Apply the chain rule for $u$ as a function of $x$.

3. Set up the expression for the derivative, but do not compute the final answer yet.

Try solving on your own before revealing the answer!

Final Answer:

$\frac{d}{dx}\cot^{-1}(u) = -\frac{1}{1+u^2} \cdot \frac{du}{dx}$

The chain rule is applied to the standard derivative of the inverse cotangent function.

Q11. What is the derivative of $\cos^{-1}(u)$ with respect to $x$?

Background

Topic: Derivatives of Inverse Trigonometric Functions

This question tests your ability to differentiate the inverse cosine function, where $u$ is a differentiable function of $x$.

Key Terms and Formulas

• $\cos^{-1}(u)$: Inverse cosine function

• Chain Rule

Step-by-Step Guidance

1. Recall the basic derivative: $\frac{d}{du}\cos^{-1}(u) = -\frac{1}{\sqrt{1-u^2}}$.

2. Apply the chain rule for $u$ as a function of $x$.

3. Set up the expression for the derivative, but do not compute the final answer yet.

Try solving on your own before revealing the answer!

Final Answer:

$\frac{d}{dx}\cos^{-1}(u) = -\frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$

The chain rule is applied to the standard derivative of the inverse cosine function.

Q12. What is the derivative of $\sec^{-1}(u)$ with respect to $x$?

Background

Topic: Derivatives of Inverse Trigonometric Functions

This question tests your ability to differentiate the inverse secant function, where $u$ is a differentiable function of $x$.

Key Terms and Formulas

• $\sec^{-1}(u)$: Inverse secant function

• Chain Rule

Step-by-Step Guidance

1. Recall the basic derivative: $\frac{d}{du}\sec^{-1}(u) = \frac{1}{|u|\sqrt{u^2-1}}$.

2. Apply the chain rule for $u$ as a function of $x$.

3. Set up the expression for the derivative, but do not compute the final answer yet.

Try solving on your own before revealing the answer!

Final Answer:

$\frac{d}{dx}\sec^{-1}(u) = \frac{1}{|u|\sqrt{u^2-1}} \cdot \frac{du}{dx}$

The chain rule is applied to the standard derivative of the inverse secant function.

Q13. What is the derivative of $a^u$ with respect to $x$?

Background

Topic: Derivatives of Exponential Functions

This question tests your ability to differentiate exponential functions with base $a$, where $u$ is a differentiable function of $x$.

Key Terms and Formulas

• $a^u$: Exponential function with base $a$

• Chain Rule

Step-by-Step Guidance

1. Recall the basic derivative: $\frac{d}{du}a^u = a^u \ln(a)$.

2. Apply the chain rule for $u$ as a function of $x$.

3. Set up the expression for the derivative, but do not compute the final answer yet.

Try solving on your own before revealing the answer!

Final Answer:

$\frac{d}{dx}a^u = a^u \ln(a) \cdot \frac{du}{dx}$

The chain rule multiplies the derivative of the outer function by the derivative of the inner function.

Q14. What is the derivative of $\tan(u)$ with respect to $x$?

Background

Topic: Derivatives of Trigonometric Functions

This question tests your ability to differentiate the tangent function, where $u$ is a differentiable function of $x$.

Key Terms and Formulas

• $\tan(u)$: Tangent function

• Chain Rule

Step-by-Step Guidance

1. Recall the basic derivative: $\frac{d}{du}\tan(u) = \sec^2(u)$.

2. Apply the chain rule for $u$ as a function of $x$.

3. Set up the expression for the derivative, but do not compute the final answer yet.

Try solving on your own before revealing the answer!

Final Answer:

$\frac{d}{dx}\tan(u) = \sec^2(u) \cdot \frac{du}{dx}$

The chain rule multiplies the derivative of the outer function by the derivative of the inner function.

Q15. What is the derivative of $\ln(u)$ with respect to $x$?

Background

Topic: Derivatives of Logarithmic Functions

This question tests your ability to differentiate the natural logarithm function, where $u$ is a differentiable function of $x$.

Key Terms and Formulas

• $\ln(u)$: Natural logarithm function

• Chain Rule

Step-by-Step Guidance

1. Recall the basic derivative: $\frac{d}{du}\ln(u) = \frac{1}{u}$.

2. Apply the chain rule for $u$ as a function of $x$.

3. Set up the expression for the derivative, but do not compute the final answer yet.

Try solving on your own before revealing the answer!

Final Answer:

$\frac{d}{dx}\ln(u) = \frac{1}{u} \cdot \frac{du}{dx}$

The chain rule multiplies the derivative of the outer function by the derivative of the inner function.

Q16. What is the derivative of $\sin^{-1}(u)$ with respect to $x$?

Background

Topic: Derivatives of Inverse Trigonometric Functions

This question tests your ability to differentiate the inverse sine function, where $u$ is a differentiable function of $x$.

Key Terms and Formulas

• $\sin^{-1}(u)$: Inverse sine function

• Chain Rule

Step-by-Step Guidance

1. Recall the basic derivative: $\frac{d}{du}\sin^{-1}(u) = \frac{1}{\sqrt{1-u^2}}$.

2. Apply the chain rule for $u$ as a function of $x$.

3. Set up the expression for the derivative, but do not compute the final answer yet.

Try solving on your own before revealing the answer!

Final Answer:

$\frac{d}{dx}\sin^{-1}(u) = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$

The chain rule is applied to the standard derivative of the inverse sine function.

Q17. Find $\frac{d}{dx}\left(\frac{u}{v}\right)$ where $u$ and $v$ are differentiable functions of $x$.

Background

Topic: Quotient Rule for Derivatives

This question tests your ability to use the quotient rule to differentiate a ratio of two functions.

Key Terms and Formulas

• Quotient Rule: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$

Step-by-Step Guidance

1. Identify $u$ and $v$ as differentiable functions of $x$.

2. Recall the quotient rule formula for derivatives.

3. Set up the expression using the formula, but do not compute the final answer yet.

Try solving on your own before revealing the answer!

Final Answer:

$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$

This is the standard quotient rule for differentiating a ratio of two functions.

Q18. Find $\frac{d}{dx}(uv)$ where $u$ and $v$ are differentiable functions of $x$.

Background

Topic: Product Rule for Derivatives

This question tests your ability to use the product rule to differentiate the product of two functions.

Key Terms and Formulas

• Product Rule: $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$

Step-by-Step Guidance

1. Identify $u$ and $v$ as differentiable functions of $x$.

2. Recall the product rule formula for derivatives.

3. Set up the expression using the formula, but do not compute the final answer yet.

Try solving on your own before revealing the answer!

Final Answer:

$\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$

This is the standard product rule for differentiating the product of two functions.

Q19. Find $\frac{dy}{dx}$ where $y = f^{-1}(x)$.

Background

Topic: Derivative of Inverse Functions

This question tests your ability to find the derivative of an inverse function.

Key Terms and Formulas

• Inverse Function: $y = f^{-1}(x)$

• Derivative Formula: $\frac{dy}{dx} = \frac{1}{f'(y)}$

Step-by-Step Guidance

1. Recall the formula for the derivative of an inverse function.

2. Set up the expression using the formula, but do not compute the final answer yet.

Try solving on your own before revealing the answer!

Final Answer:

$\frac{dy}{dx} = \frac{1}{f'(y)}$

This formula gives the derivative of the inverse function in terms of the derivative of the original function evaluated at $y$.

Q20. Find the derivative of $f(g(x))$ with respect to $x$.

Background

Topic: Chain Rule for Derivatives

This question tests your ability to use the chain rule to differentiate a composite function.

Key Terms and Formulas

• Chain Rule: $\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$

Step-by-Step Guidance

1. Identify $f$ and $g$ as differentiable functions.

2. Recall the chain rule formula for derivatives.

3. Set up the expression using the formula, but do not compute the final answer yet.

Try solving on your own before revealing the answer!

Final Answer:

$\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$

This is the standard chain rule for differentiating a composite function.